two physiscs students are on the top of a high rise building one student dropped

Question

two physiscs students are on the top of a high rise building one student dropped a tennis ball, 3 seconds latter student two throw a stone downward with a velocity of 10 m/s. the stone and the tennis ball reached the ground of the same instant.

a) How high is the building?

b) what is the time it takes for the stone and the ball to reach the ground?

c) what is the velocity of the stone & the ball when it reach the ground?

d) where is the stone & the ball at 5 seconds after the ball is dropped.

Explanation / Answer

Let h is the height of the building.

Let v be the velocity of the stone when reaches the ground.

Thus Apply second equation of motion

h = u(t-3) + 0.5*g*(t-3)^2 = 10(t-3) + 4.9(t-3)^2 ------(1)



Because  time taken by the stone = (t-3) sec

Therefore for ball initial velocity u = 0 m/sec

Therefore apply second equation of motion

h = 0 + 4.9t^2 = 4.9t^2

Put this value in eq(1)

4.9t^2 = 10(t-3) + 4.9(t-3)^2

0 = 10t - 30 + 44.1 - 29.4t

Thus  19.1t = 14.1

Thus t = 0.7 sec

Thus height of the building h = 4.9*0.7^2 = 2.40 m

Let velocity of the ball when reaches the ground = v m/sec

This Apply first equation of motion

v = 0 +9.8*0.7 = 6.86 m/sec

Let the velocity of the stone when hit the ground = V m/sec

Thus apply first equation of motion

V = 10 +9.8*0.7 = 16.86 m/sec

Distance travel by the ball after 5 second = 4.9*5^2 = 122.5 m

Distance travel by the stone in (5-3)sec = 10*2 + 4.9*4  = 39.6 m

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