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What are 3 Examples of Projectile Motion?
Definition: A body thrown upwards in any direction from the earth’s surface or a point close to it, is called a projectile.
Common examples of projectiles are-
- a javelin thrown by an athlete,
- a bullet fired from a rifle,
- an object dropped from an aeroplane,
- a jet of water coming out from the side hole of a vessel,
- a stone thrown from the top of a hill or a tower,
- a rocket after its fuel is exhausted.
The path traced out by a projectile is called its trajectory. The motion of a projectile is two-dimensional as it is always confined on a vertical plane.
To study the projectile motion the horizontal direction is taken along the x-axis and the vertical direction is taken along the y-axis. The only force acting on the projectile is gravitational force.
So the projectile has acceleration in the y direction which is equal to the acceleration due to gravity. The force acting on the projectile is zero along the x-axis. So the horizontal acceleration is zero. This means, projectile motion is a combination of horizontal motion with constant velocity and vertical motion with uniform acceleration.
Let a body be projected from a point O with velocity u making an angle α with the horizontal. The body reaches B following the path OAB through the highest point A. Points B and O lie on the same horizontal plane [Fig.].
O is the point of projection, u the velocity of projection, α the angle of projection. OB the horizontal range; the time taken for travelling the path OAB is called the time of flight. Velocity of projection u has a horizontal component = \(u_{x_0}\) = ucosα and a vertical component = \(u_{y_0}\) = usinα. As acceleration due to gravity (g) acts vertically downward, the velocity component usinα gradually changes. Hence, the vertical motion of the body is a motion under gravity [see the section 1.10 in the chapter One-dimensional Motion]. But since g has no component along the horizontal direction, the acceleration or deceleration of the horizontal component of motion is zero. Hence, the motion of the body along the horizontal direction is uniform, i.e., horizontal velocity = \(u_{x_0}\) = ucosα = constant. It is convenient to use the vertical motion and the horizontal motion separately in discussions related to projectile motion.
Principle of physical independence at motions: In the absence of air resistance the motion of a projectile is considered as the combination of the following two independent motions.
- Motion along horizontal direction with uniform velocity.
- Motion along vertical direction under gravity i.e., uniform acceleration equal to g.
The two motions of a projectile along horizontal and vertical directions are independent of each other. This is called the principle of physical independence of motions.
Key points:
i) A projectile returns to the ground at the same angle and with the same velocity with which it is projected.
ii) When a projectile is at the highest point of its trajectory
1. it possesses velocity only along horizontal,
2. the velocity and acceleration of the projectile are perpendicular to each other.
Equations for projectile motion: Let the time taken by the projectile to reach the point P [Fig.] be t.
For horizontal motion of the projectile we have,
vx = \(u_{x_0}\) + axt
or, vx = ucosα [∵ ax = 0] …. (1)
where vx is the horizontal component of the velocity of projectile after time t.
Again, x = \(u_{x_0} t\) + \(\frac{1}{2} a_x t^2\)
or, x = \(u_{x_0} t\) =ucosαt ……. (2)
From this equation we get the horizontal position of the projectile after time t.
For vertical motion of the projectile,
ay = -g ∴ uy = \(u_{y_0}\) + \(a_y t\)
[vx is the vertical component of velocity of the projectile after time t]
or, vy = \(u_{y_0}\) – gt ……… (3)
Again, \(v_y^2\) = \(u_{y_0}^2\) – 2gy
and y = \(u_{y_0} t\) + \(\frac{1}{2} a_y t^2\) = \(u_{y_0} t\) – \(\frac{1}{2} g t^2\) …… (4)
From equation (4), we get the vertical position of the projectile after time t.
Maximum height: Maximum height of a projectile is the maximum vertical distance attained by the projectile above the horizontal plane of projection. It is denoted by H.
To calculate the height of the projectile, consider only its vertical motion. The vertical component of the velocity of projection = u sin α and velocity at the highest point = 0.
If H = maximum height,
0 = (usinα)2 – 2gH or, H = \(\frac{u^2 \sin ^2 \alpha}{2 g}\) ….. (5)
It should be noted that the body attains maximum height by reaching a point which is not exactly above the point O. Because of the horizontal component of motion, the body has a horizontal displacement as well. In Fig., the highest point on the trajectory of the projectile is the point A and maximum height (H) = AC.
Time of flight: It is the time taken by the projectile from the instant it is projected till it reaches a point in the horizontal plane of its projection. As shown in Fig. the total time taken to reach point B from point O is the time of flight.
Let the time taken to reach the maximum height be T1.
Hence, 0 = usinα – gT1 or, T1 = \(\frac{u \sin \alpha}{g}\) …… (6)
On reaching the highest point, the body starts descending again. At the same time, the body covers a horizontal path due to its horizontal component of motion. Finally, the body reaches B. Thus, the body follows the trajectory OAB.
The height of B with respect to the point of projection, O is zero, i.e., the vertical displacement of the object is zero. Hence, if T is the time required to cover the path QAB,
0 = usinαᐧT – \(\frac{1}{2} g T^2\) [from equation(4)]
2usinα = gT or, T = \(\frac{2 u \sin \alpha}{g}\) …… (7)
Comparing equations (6) and (7),
2T1 = \(\frac{2 u \sin \alpha}{g}\) = T or, T1 = \(\frac{T}{2}\)
So, the time required to reach the maximum height is equal to half the time of flight.
Horizontal range: Let the horizontal range be OB = R. The horizontal distance traversed by the body in time T is R. Since the body crosses the distance R horizontally with a uniform velocity ucosα,
R = ucosα ᐧ T = ucosαᐧ\(\frac{2 u \sin \alpha}{g}\)
= \(\frac{2 u^2 \sin \alpha \cos \alpha}{g}\) = \(\frac{u^2}{g} \sin 2 \alpha\) ….. (8)
The value of R is maximum when sin2α is maximum. Therefore, the condition for covering the maximum horizontal distance for a particular initial velocity u is
sin2α = 1 = sin90° or, α = 45°
∴ Rmax= Maximum range = \(\frac{u^2}{g}\) … (9)
To send a projectile to the maximum possible distance, it has to be thrown at an angle of 45° with the horizontal. This is why sportsmen try to throw javelin or discus at 45°.
It should be noted that if the angle of projection is (\(\frac{\pi}{2}\) – α) instead of α, the horizontal range remains the same for a particular velocity of projection, since,
R = \(\frac{u^2}{g}\)sin{\(\left\{2\left(\frac{\pi}{2}-\alpha\right)\right\}\) = \(\frac{u^2}{g}\)sin(π – 2α) = \(\frac{u^2}{g}\)sin2α = R
Locus of a projectile: Let us take a reference frame such that, its positive y -axis extends vertically upwards, and the positive x -axis extends horizontally in the direction of the horizontal component of the velocity of the projectile. The origin is taken as the point of projection [Fig.]
Let the point reached by the body from O in time t be P and the coordinates of P be (x, y).
y = usinα ᐧ t – \(\frac{1}{2} g t^2\) …… (10)
Again, as the horizontal velocity = ucosα = constant,
x = ucosα ᐧ t
or, t = \(\frac{x}{u \cos \alpha}\) …….. (11)
From (10) and (11),
y = usinα\(\frac{x}{u \cos \alpha}\) – \(\frac{1}{2} g \frac{x^2}{u^2 \cos ^2 \alpha}\)
or, y = xtanα – \(\frac{g}{2 u^2 \cos ^2 \alpha} x^2\)
This equation is the locus of the projectile.
The equation is of the type y = ax + bx2 which is the equation of a parabola. Hence, the trajectory of a projectile is parabolic.
Numerical Examples
Example 1.
A body is projected with a velocity 20 m ᐧ s-1, making an angle of 45° with the horizontal. Calculate—
(i) the time taken to reach the ground [g = 10 m ᐧ s-2],
(ii) the maximum height it can attain and
(iii) horizontal range.
Solution:
The vertical and horizontal components of the velocity of 20 m ᐧ s-1 are
uH = 20 cos45° = 20 × \(\frac{1}{\sqrt{2}}\) = 10\(\sqrt{2}\) m ᐧ s-1
and uV = 20 sin45° = 20 × \(\frac{1}{\sqrt{2}}\) = 10\(\sqrt{2}\) m ᐧ s-1
i) Let the total time of flight of the body be t. Considering the vertical motion of the body, we get from the equation h = ut – \(\frac{1}{2}\)gt2,
0 = 10\(\sqrt{2}\)t – \(\frac{1}{2}\) ᐧ 10 ᐧ t2 [as h = 0]
or 5t2 = 10\(\sqrt{2}\)t
∴ Total time of flight,
t = \(\frac{10 \sqrt{2}}{5}\) = 2\(\sqrt{2}\) = 2 × 1.414 = 2.828 s.
ii) Let the maximum height attained be h . Vertical velocity at the maximum height = 0.
Considering the vertical motion of the body, we get from the equation u2 = u2 – 2gh,
0 = (10\(\sqrt{2}\))2 – 2 ᐧ 10 ᐧ h ∴ h = 10 m.
iii) Let the distance from the point of projection to the point at the ground where the body touches be x.
By considering the horizontal motion of the body, we get,
x = UH × t = 10\(\sqrt{2}\) × \(\sqrt{2}\) = 40m.
∴ The horizontal range = 40 m.
Example 2.
A plane is flying horizontally at a height of 1960 m at 600 km ᐧ h-1 with respect to the ground. On reaching a point directly above A, the plane drops an object which reaches the ground at B. Find the distance AB.
Solution:
Let the point directly above A from where the object is dropped be O [Fig.].
Therefore, OA = 1960 m. Let the 600 time taken by the object to hit the ground at B be t.
Since the plane is flying horizontally, the initial vertical velocity of the object = 0 and vertical (downward) displacement = 1960 m.
Considering the vertical motion of the object, we get from the equation h = ut + \(\frac{1}{2}\)gt2,
1960 = 0 × t + \(\frac{1}{2}\) × 9.8 × t2
or, t2 = \(\frac{2 \times 1960}{9.8}\) = 400 or, t = 20 s
The initial horizontal velocity of the body
= 600 km ᐧ h-1 = \(\frac{600 \times 1000}{60 \times 60}\) m ᐧ s-1 = \(\frac{500}{3}\) m ᐧ s-1
So, the object moves at a uniform velocity of \(\frac{500}{3}\) m ᐧ s-1 in
the horizontal direction. Hence,
AB = \(\frac{500}{3}\) × 20 m = \(\frac{500 \times 20}{3 \times 1000}\)km = \(\frac{10}{3}\) = 3.33 km.
Example 3.
A particle falls from rest from the highest point of a vertical circle of radius r, along a chord without any friction. Show that the time taken by the particle to come down is independent of the chords length. Find the time in terms of r and g.
Solution:
Let the chord along which the particle falls be CD [Fig.]. CD makes an angle θ with the vertical diameter as shown. CD = 2r cosθ. The component of acceleration due to gravity along CD = g cosθ. Let C the time taken by this particle to fall from rest along CD be t.
Hence, from the equation s = ur + \(\frac{1}{2}\)at2,
2r cosθ = 0 + \(\frac{1}{2}\)gcosθ ᐧ t2
or, t2 = \(\frac{4 r}{g}\) or, t = 2\(\sqrt{\frac{r}{g}}\).
The time is independent of θ, and hence on the length of the chord CD.
Example 4.
At what angle with respect to the horizontal, should a projectile be thrown with a velocity of 19.6 m ᐧ s-1, to just clear a wall 14.7 m high, at a distance of 19.6 m?
Solution:
Let the angle of projection be θ [Fig.]. Hence horizontal component of velocity = 19.6 cosθ m ᐧ s-1 and its vertical component
= 19.6 sinθ m ᐧ s-1
Let the time after which the projectile crosses the wall be t.
Considering horizontal motion,
19.6 = 19.6 cosθ × t
or, t = secθ
For the vertical motion,
14.7 = 19.6 sinθ × t – \(\frac{1}{2}\) × 9.8 × t2
or, 14.7 = 19.6 sinθ × secθ – 4.9 sec2θ
or, 3 = 4 tanθ – (1 + tan2θ) or, tan2θ – 4tanθ + 4 = 0
or, (tanθ – 2)2 = 0 or, tanθ = 2
∴ θ = tan-12 = 63.4°.
Example 5.
A block of ice is sliding down the sloping roof of a house and the angle of Inclination of the roof with the horizontal is 30°. The maximum and minimum heights of the roof from the ground are 8.1 m and 5.6 m. How far from the starting point, measured horizontally, does the block land? [Ignore friction]
Solution:
Let the highest point of the roof be A and the lowest point be P as shown in Fig.
∴ AC = 8.1 m ; PD = 5.6 m
∴ AB = AC – BC = AC – PD = 8.1 – 5.6 = 2.5 m
AP = \(\frac{A B}{\sin 30^{\circ}}\) = \(\frac{2.5}{\frac{1}{2}}\) = 5 m
Let the velocity of the block at P be v.
Considering the motion of the block from A to P,
v2 = 2g sin30° × AP = 2 × 9.8 × \(\frac{1}{2}\) × 5 = 49
or, v = 7 m ᐧ s-1.
The horizontal and vertical components of the velocity at P are v cos30° = \(\frac{7 \sqrt{3}}{2}\)m ᐧ s-1 and vsin30° = \(\frac{7}{2}\)m ᐧ s-1 respectively.
Let the total time taken by the block to come from P to E be t. Considering the vertical motion of the block,
Example 6.
The equation of the trajectory of a projectile on a vertical plane is y = ax – bx2 where a and b are constants, and x and y respectively are the horizontal distances of the projectile from the point of projection. Find out the maximum height attained by the projectile, and also the angle of projection with respect to the horizontal.
Solution:
Let, u = velocity of projection; α = angle of projection.
The velocity ucosci in the horizontal direction is uniform.
So, in time t,
x = ucosα ᐧ t or, t = \(\frac{x}{u \cos \alpha}\)
The velocity usinα in the vertically upward direction is under a uniform retardation -g, where g is the acceleration due to gravity Then, in time t,
y = usinα ᐧ t – \(\frac{1}{2}\)gt2 = usinα ᐧ \(\frac{x}{u \cos \alpha}\) – \(\frac{1}{2} g \frac{x^2}{u^2 \cos ^2 \alpha}\)
or, y = xtanα – \(\frac{g}{2 u^2 \cos ^2 \alpha} x^2\)
Comparing with the given equation y = ax – bx2, we get
- a = tan θ, or angle of projection, θ = tan-1 a.
- b = \(\frac{g}{2 u^2 \cos ^2 \alpha}\) or, u2 = \(\frac{g}{2 b \cos ^2 \alpha}\)
At maximum height H, the velocity of the projectile is zero. Considering vertical motion, we have
0 = (usinα)2 – 2gH
or H = \(\frac{u^2 \sin ^2 \alpha}{2 g}\) = \(\frac{g}{2 b \cos ^2 \alpha} \cdot \frac{\sin ^2 \alpha}{2 g}\)
= \(\frac{\tan ^2 \alpha}{4 b}\) = \(\frac{a^2}{4 b}\)
Example 7.
A gun fires at an angle 30° with the horizontal and hits a target at a distance of 3 km. Can another tar get at a distance of 5 km be hit by changing the angle of projection but keeping the velocity of projection unchanged?
Solution:
Horizontal range, R = \(\frac{u^2 \sin 2 \alpha}{g}\)
In the first case, 3 = \(\frac{u^2 \sin \left(2 \times 30^{\circ}\right)}{g}\) = \(\frac{u^2}{g} \frac{\sqrt{3}}{2}\)
or, \(\frac{u^2}{g}\) = 2\(\sqrt{3}\)
If the velocity of projection is unchanged, the maximum horizontal range for α = 45°, is
Rmax = \(\frac{u^2 \sin \left(2 \times 45^{\circ}\right)}{g}\) = \(\frac{u^2}{g}\) = 2\(\sqrt{3}\)
= 2 × 1.732 = 3.464 km
So, a target at a distance of 5 km cannot be hit.
Example 8.
A gun is kept on a horizontal road and is used to hit a running car. The uniform speed of the car is
72 km/h. At the instant of firing at an angle of 45° with the horizontal, the car is at a distance of 500 m from the gun. Find out the distance between the gun and the car at the instant of hitting. Given, g = 10 m/s2.
Solution:
Velocity of the car, v = 72km/h = 20 m/s;
if it is hit after a time t s, then its displacement = 20 t m
∴ Distance between the gun and the car at that instant,
D = 500 + 20t m.
If u be the initial velocity of the bullet, then its horizontal range,
R = \(\frac{u^2 \sin \left(2 \times 45^{\circ}\right)}{g}\) = \(\frac{u^2}{g}\)
∴ \(\frac{u^2}{g}\) = 500 + 20t ……. (1)
Considering the vertical motion of the bullet in time t, we have
Example 9.
The initial velocity of a projectile is \((\hat{i}+2 \hat{j})\) m/s, where \(\hat{i}\) and \(\hat{j}\) are unit vectors along the horizontal and vertical directions respectively. Find out the locus of the projectile, taking g = 10 m/s2. [JEE Main ‘13]
Solution:
Horizontal and vertical components of initial velocity are, respectively,
ux = 1 m/s and uy = 2 m/s
Let, at t = 0, the initial coordinates of the projectile are (0, 0);at time t, these are (x, y).
So, x = uxt = tm;
y = uyt – \(\frac{1}{2}\)gt2 = (2t – 5t2) = 2x – 5x2
∴ y = 2x – 5x2 is the locus of the projectile.
Example 10.
Two objects are thrown simultaneously from the same point with the same initial velocity at angles of projection α and β respectively. If they reach the top and the bottom of a tower simultaneously, then prove that tanα – tanβ = tanθ where, θ = angle of elevation of the tower from the point of projection.
Solution:
Horizontal range of the 2nd projectile [Fig.]
Time of flight of the 2nd projectile,
t = \(\frac{2 u \sin \beta}{g}\)
= time taken by the 1st projectile from O to A
For the horizontal motion of the 1st projectile,
OB = x = ucosα ᐧ t = ucosα ᐧ \(\frac{2 u \sin \beta}{g}\)
= \(\frac{2 u^2}{g}\)sinβcosα …… (2)
Comparing (1) and (2), we have
cosα = cosβ …… (3)
For the vertical motion of the 1st projectile,
AB = y = usinα ᐧ t – \(\frac{1}{2} g t^2\)
= usinα\(\frac{2 u \sin \beta}{g}\) – \(\frac{1}{2} g\left(\frac{2 u \sin \beta}{g}\right)^2\)
= \(\frac{2 u^2}{g}\) (sinαsinβ – sin2β) ….. (4)
Dividing (4) by (1), we have