Physics Topics can help us understand the behavior of the natural world around us.
Is vertical Motion Downward Due to Gravity
Acceleration due to gravity: When an object is released from a certain height above the earth’s surface, it moves with vertically downward acceleration. Again, when an object is thrown vertically upwards from the ground, it moves up with a deceleration. An upward deceleration is equivalent to a downward acceleration. Actually, the acceleration is always downwards, and its magnitude is the same for both downward and upward motions.
So the equations of vertical motion are,
v = u + at …… (1)
h = ut + \(\frac{1}{2}\)at2 ………. (2)
and v2 = u2 + 2ah ……. (3)
Here, initial velocity is u, velocity after time t is v, vertical displacement in time t is h, and acceleration is a.
This acceleration a is called the acceleration due to gravity or free fall acceleration and is represented by the letter g.
The direction of g is always vertically downwards.
If the downward direction is taken as positive, i.e., a = g, then we get the following equations of motion:
v = u + gt ….. (4)
h = ut + \(\frac{1}{2}\)gt2 ……… (5)
and v2 = u2 + 2gh ….. (6)
If the upward direction is taken as positive, i.e., a = -g, then we get the following equations:
v = u – gt …. (7)
h = ut – \(\frac{1}{2}\)gt2 ….. (8)
and v2 = u2 – 2gh …. (9)
Maximum height attained: When a body is thrown vertically upwards with a velocity u, it momentarily comes to rest on attaining the maximum height and then again starts falling vertically downwards.
Hence, at maximum height H, the velocity of the body v = 0.
∴ 0 = u2 – 2gH [from equation (9)]
or, H = \(\frac{u^2}{2 g}\) ….. (10)
Time to reach the maximum height: Let the time required to reach the maximum height be T.
∴ 0 = u – gT [from equation (7)]
or, T = \(\frac{u}{g}\) …… (11)
Time taken total from the maximum height: If T1 is the time taken by the body to fall from the maximum height to the initial position then, using equation (5),
H = \(\frac{1}{2} g T_1^2\) [as at maximum height, u = 0]
or, \(T_1^2\) = \(\frac{2 H}{g}\) = \(\frac{2}{g} \cdot \frac{u^2}{2 g}\) = \(\frac{u^2}{g^2}\) ∴ T1 = \(\frac{u}{g}\) ….. (12)
Hence, the time to reach the maximum height is equal to the time to return to the starting point.
Time of flight: It is the total time required for upward and downward motions,
T’ = T + T1 = \(\frac{u}{g}\) + \(\frac{u}{g}\) = \(\frac{2 u}{g}\)
Alternative method:
After completion of the upward and downward motions, the displacement becomes zero. Thus, using equation (8),
0 = uT’ – \(\frac{1}{2}\)T’2 or, u = \(\frac{1}{2}\)gT’ or, T’ = \(\frac{2 u}{g}\).
Time taken to reach a certain height: Let the object reach a height h at time t. The initial upward velocity = u.
∴ h = ut – \(\frac{1}{2} g t^2\)
or, \(\frac{1}{2} g t^2\) – ut + h = 0
or, t = \(\frac{u}{g} \pm \frac{\sqrt{u^2-2 g h}}{g}\)
From this equation, two different values of t are obtained. This is because the object crosses the point at a height h twice, first during the upward motion and then during the downward motion.
Velocity at any héight: From the equation v2 = u2 – 2gh we get, v = ±\(\sqrt{u^2-2 g h}\).
Positive (+) sign is applicable for upward motion and negative (-) sign is applicable for downward motion. So, an object crosses any point with the same magnitude of velocity in its upward and downward motions.
Velocity of projection and velocity of return: Let u = velocity of projection, v = velocity of return and total time of flight = T’.
∴ From equation (7), we can write,
v = ut – gT’ = u – gᐧ\(\frac{2 u}{g}\) = -u[∵ T’ = \(\frac{2 u}{g}\)]
Hence, the upward velocity of projection is equal in magnitude to the downward velocity with which an object hits the ground.
Numerical Examples
Example 1.
A stone is dropped from a height of 19.6 m. What is the time taken by the stone to travel the last metre of the path? [HS(XI) ’06]
Solution:
In this case u = 0
∴ h = \(\frac{1}{2} g t^2\) or, t = \(\sqrt{\frac{2 h}{g}}\)
Let t1 and t2 be the time taken by the stone to travel (19.6 – 1) = 18.6 m and 19.6 m, respectively.
∴ t1 = \(\sqrt{\frac{2 \times 18.6}{9.8}}\) and t2 = \(\sqrt{\frac{2 \times 19.6}{9.8}}\)
or, t1 = 1.948 s and t2 = 2 s
Hence, the time taken to travel the last metre, t2 – t1 = 2 – 1.948 = 0.052 s.
Example 2.
An object is thrown vertically upwards with an initial velocity of 40m s-1.
(i) How long will the object move upwards?
(ii) What will be the maximum height attained?
(iii) How much time will it take to reach the ground?
(iv) When will the object be at a height of 25 m from the ground?
(v) What will be its velocity after 2 s ? [g = 9.8 m ᐧ s-2]
Solution:
i) Let the time taken for upward motion be t s. At max-imum height its velocity is zero. From equation v = u – gt, we get,
0 = 40 – 9.8 × t or, t = 4.1s.
ii) Let the maximum height attained be h.
From equation v2 = u2 – 2gh, we get,
0 = (40)2 – 2 × 9.8h or, h = 81.6 m.
iii) Let the time taken to reach the ground be t1 starting from the time of projection. Considering both upward and downward motions of the body and using the equation h = ut – \(\frac{1}{2}\)gt2, we get,
0 = 40t1 – \(\frac{1}{2}\) × 9.8 × \(t_1^2\) [total displacement is zero in this case]
∴ t1 = 8.2 s.
iv) Let the time after which the body is at a height of 25 m be x.
Hence, from h = ut – \(\frac{1}{2}\)gt2 we get,
25 = 40x – \(\frac{1}{2}\) ᐧ 9.8x2 or, 49x2 – 400x + 250 = 0
or, x = 0.682 s and x = 7.481 s
Two values of x signify that the object will be at a height of 25 m twice during its flight, once (x = 0.682 s) while moving upwards and the next (x = 7.481 s) during its downward motion.
v) Let the velocity acquired 2 s after the projection be v.
∴ v = 40 – 9.8 × 2 = 20.4 m ᐧ s-1.
Example 3.
A ball falls freely on a perfectly elastic plate from a height of 3 m. At the instant t = 0, the velocity of the ball is zero. Draw velocity-time graph for the motion of the ball, [g = 9.8 m ᐧ s-2]
Solution:
In this case, h = \(\frac{1}{2}\)gt2
The time taken by the ball in falling through 3 m is given by
t = \(\sqrt{\frac{2 h}{g}}\) = \(\sqrt{\frac{2 \times 3}{9.8}}\) = 0.783 s
According to the equation, v2 = 2gh, the velocity of the ball just before striking the plate is given by
Since the plate is perfectly elastic, the ball after striking it will rebound with the same velocity (7.67 m ᐧ s-1) and its velocity will become zero after the same time (0.78 s). This motion will be repeated again and again as shown in the above graph.
Example 4.
A body is thrown vertically upwards. After attaining half of its maximum height its velocity becomes 14 m ᐧ s-1.
(i) How high will the body rise?
(ii) What will be the velocity of the body 1 s and 3 s after its projection?
(iii) What is the average velocity of the body in the first half second?
Solution:
i) Let the velocity of projection be u.
Hence, maximum height attained by the body,
h = \(\frac{u^2}{2 g}\) [using v2 = u2 – 2gh] or, u2 = 2 gh
For half the maximum height, i.e., \(\frac{h}{2}\), we get
(14)2 = u2 – 2g\(\frac{h}{2}\) = 2 gh – \(\frac{1}{2}\)(2gh) = gh
∴ h = 20 m [where g = 9.8 m ᐧ s-2]
∴ The body will rise up to a height of 20 m.
ii) Here, the velocity of projection,
u = \(\sqrt{2 g h}\) = \(\sqrt{2 \times 9.8 \times 20}\) = 19.8 m ᐧ s-1
∴ Velocity of the body 1 s after projection is,
v1 = u – g ᐧ 1 = 19.8 – 9.8 = 10 m ᐧ s-1.
Velocity 3 s after projection,
v2 = u – g × 3 = 19.8 – 9.8 × 3 = -9.6 m ᐧ s-1
(Negative sign indicates downward motion of the body.)
iii) Velocity after \(\frac{1}{2}\)s,
v’ = u – g × \(\frac{1}{2}\) = 19.8 – 9.8 × \(\frac{1}{2}\) = 14.9 m ᐧ s-1
Hence, average velocity during the given period
= \(\frac{u+v^{\prime}}{2}\) = \(\frac{(19.8+14.9)}{2}\) = 17.35 m ᐧ s-1
Example 5.
A piece of stone was dropped from a stationary balloon. The stone covered 13.9 m during the last \(\frac{1}{7}\) s of its descent. Find the height of the balloon and the velocity of the stone when it strikes the ground. [g = 9.8 m ᐧ s-2]
Solution:
Let h = height of the balloon, t = total time of fall of the stone, h’ = downward displacement in time (t – \(\frac{1}{7}\)) s.
Hence, h – h’ = \(\frac{1}{2} g t^2\) – \(\frac{1}{2} g\)(t – \(\frac{1}{7}\))2
or, 13.9 = \(\frac{1}{2}\) × 9.8 × t2 – \(\frac{1}{2}\) × 9.8(t – \(\frac{1}{7}\))2
or, 13.9 = 4.9t2 – 4.9t2 + 1.4t – 0.1
or, 1.4t = 14 or, t = 10
∴ Height of the balloon,
h = \(\frac{1}{2}\)gt2 = \(\frac{1}{2}\) × 9.8 × (10)2 = 490 m
Velocity of the stone when it strikes the ground is,
v = u + gt = 0 + 9.8 × 10 = 98 m ᐧ s-1.
Example 6.
A stone is dropped from the top of a tower 400 m high. At the same time another stone is thrown upwards from the ground with a velocity of 100 m ᐧ s-1. When and where will they meet each other? 1g = 9.8 m ᐧ s-2] [HS ‘01]
Solution:
Let the two stones meet after a time t at a distance h from the top of the tower.
Considering the downward motion of the 1st stone,
h = \(\frac{1}{2} g t^2\) = \(\frac{1}{2} \times 9.8 t^2\) ….. (1)
Considering the upward motion of the 2nd stone,
400 – h = 100t – \(\frac{1}{2} \times 9.8 t^2\) ….. (2)
From equations (1) and (2) we get,
400 – \(\frac{1}{2}\) × 9.8t2 = 100t – \(\frac{1}{2}\) × 9.8t2
or, 100t = 400 or, t = 4 s
Hence, from equation (1) we get,
h = \(\frac{1}{2}\) × 9.8 × (4)2 = \(\frac{1}{2}\) × 9.8 × 16 = 78.4 m
Hence, the two stones meet at 78.4 m below the top of the tower after 4 s.
Example 7.
A stone is dropped from the top of a vertical pillar. When the stone has fallen through a height x, another stone is dropped from height y below the top of the pillar. Both the stones touch the ground at the same time. Prove that the height of the pillar should be \(\frac{(x+y)^2}{4 x}\).
Solution:
Let the height of the pifiar be h and the velocity of the stone at x below the top of the pillar be v.
∴ v2 = 2gx or, v = \(\sqrt{2 g x}\) …. (1)
Let the first stone take t s to cover the distance (h – x).
∴ h – x = vt + \(\frac{1}{2}\)gt2….. (2)
According to the problem, the second stone is dropped from a height of (h – y) and this stone takes time t to cross that distance.
∴ h – y = \(\frac{1}{2}\)gt2 …. (3)
From equations (2) and (3) we get,
Example 8.
A, B, C and D are four polntš on a vertical line such that AB = BC = CD. A body is allowed to fall freely from A. Prove that the respective times required by the body to cross the distances AB, BC, CD should be in the ratio 1: (\(\sqrt{2}\) – 1) : (\(\sqrt{3}\) – \(\sqrt{2}\)).
Solution:
Let AB = BC = CD = x and time taken by the body to cover these distances be t1, t2 and t3 respectively.
Example 9.
A rubber ball is thrown vertically downwards from the top of a tower with an initial velocity of 14 m ᐧ s-1. A second ball is dropped is later from the same place. In 2 s the first ball reaches the ground and rebounds upwards with the same velocity. When will they collide with each other?
Solution:
Height of the tower, h = distance covered by the first ball in 2s = 14 × 2 + \(\frac{1}{2}\) × 9.8 × (2)2 = 47.6 m
[∵ h = ut + \(\frac{1}{2}\)gt2]
Velocity of the first ball just before touching the ground is
v = 14 + 9.8 × 2 = 33.6 m ᐧ s-1
Hence, its velocity just after bouncing = 33.6 m ᐧ s-1
Downward displacement of the second ball in 1 s,
x = \(\frac{1}{2}\) × 9.8 × (1)2 = 4.9m
Velocity of second ball after is = 9.8 × 1 = 9.8m ᐧ s-1
Hence, distance between the two balls, 2 s after the projection of the first ball = 47.6 – 4.9 = 42.7 m.
Let the two balls collide with each other r s after the first ball bounces off the ground.
Upward displacement of the first ball in t s,
x1 = 33.6t – \(\frac{1}{2}\) × 9.8 × t2 = 33.6t – 4.9t2
Downward displacement of the second ball in t s,
x2 = 9.8t + \(\frac{1}{2}\) × 9.8 × t2 = 9.8t + 4.9t2
Now, x1 + x2 = 42.7
or, 33.6t – 4.9t2 + 9.8t + 4.9t2 = 42.7 or, 43.4t = 42.7
∴ t = \(\frac{42.7}{43.4}\) = 0.98s.
Example 10.
A lift starts to move up with a constant acceleration of 2 m ᐧ s-2 from the earth’s surface. A piece of stone is dropped outside from the lift 4 s after the start of the lift. When will the stone reach the earth’s surface?
Solution:
Initial velocity of the lift, u = 0; acceleration, a = 2 m ᐧ s-2
Let the rise of the lift in 4 s be s and its velocity at that point be v.
Hence, from equation v = u + at we get,
v = 0 + 2 × 4 = 8 m ᐧ s-1
Also from equation s = ut + \(\frac{1}{2}\)at2 we get,
s = 0 × 4 + \(\frac{1}{2}\) × 2 × (4)2 = 16 m
∴ The stone piece was dropped with an initial upward velocity of 8 m ᐧ s-1 and was at a height of 16 m from the ground. If t is the time taken by the stone to reach the ground, then from equation h = ut + \(\frac{1}{2} g t^2\),
16 = -8t + \(\frac{1}{2}\) × 9.8t2
[∵ for the stone, u = -8 m ᐧ s-1, g = 9.8 m ᐧ s-1, h = 16 m]
or, 4.9t2 – 8t – 16 = 0
or, t = \(\frac{8 \pm \sqrt{(8)^2-4 \times 4.9 \times(-16)}}{2 \times 4.9}\) = \(\frac{8 \pm 19.4}{9.8}\)
Since time cannot be negative, t = \(\frac{8+19.4}{9.8}\) = 2.8 s.