02.3 A person throws a ball vertically upward from the top of a tall cliff, in t

Question

02.3 A person throws a ball vertically upward from the top of a tall cliff, in the positive y- nt is at yo-11.0 m. After direction, with an initial speed Vo = 19.6 m/s. The release poi the (constant) acceleration of the ball due to gravity is ay- -9.80 m/s. n this problem ignore the possible influence of air resistance on the motion of the ball. (a) How long does it take for the ball to reach its maximum height? (b) What is the maximum height of the ball before the ball starts coming back down? (c) How long does it take, from the time it is initially thrown, for the ball to reach the bottom of the cliff at y = 0.00 m? (d) What is the speed of the ball when the ball reaches the bottom of the cliff?

Explanation / Answer

a]

at the highest point, v = 0 m/s

u = 19.6 m/s

v = 0 m/s

g = 9.8 m/s2

use, v = u - gt

=> 0 = 19.6 - 9.8t

=> t = 2s

so, the time it takes for the ball to reach its maximum height is t = 2s.

b]

v2 = u2 - 2ah

=> 0 = (19.6)2 - 2(9.8)h

=> h = 19.6 m

so, the maximum height of the ball from the release point is, h = 19.6 m and the maximum height from the ground will be: H = h + yo = 30.6 m

c]

Velocity at the ground will be:

v = [2gH]1/2 = 24.489 m/s

initial velocity = velocity at the top = u = 0m/s

so,

t = [24.489/9.8] = 2.5s

so, the time it takes for the ball to reach the ground from the time it was initially thrown is:

T = 2 + 2.5 = 4.5s

d] As calculated above, the speed of the ball when it reaches the bottom will be:

v = 24.489 m/s.

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