Daphne drops a ball from the top of a tall building, of height H. The ball takes

Question

Daphne drops a ball from the top of a tall building, of height H. The ball takes a time T to reach the ground below. At the same time Daphne drops that ball, Alphonse, on the ground below, throws a second ball straight up with initial velocity V – it just reaches the top of the building. Neglect air resistance [2 points]

(a) Is the time the ball that Alphonse threw takes to reach the top of the building greater than, less than, or equal to T? Briefly explain your answer.

(b) Do the balls pass one another before, after or at T/2? Prove your answer graphically

(c) How far above the ground are the balls when they pass one another? IE are they more than, less than, or equal to H/2? Use the graph above to provide evidence to explain your answer.

Explanation / Answer

For a droped body s=ut+(1/2)gt2

u =0 (intial velocity) and t=T and a =g,s=H we get

Then substituting in the above equation T =Sqrt(2H/g)

Now for the projectd body

T =u/g ....(1)

then from the equation v2-u2 =2as

V =0 , u=u and a=-g and s=H

Then -u2 =-2gH

u =Sqrt(2gH)----(2)

Substituting equation(2) in (1) we get T =Sqrt(2H/g)

The time taken for both balls are equal to reach to the ground and to reach height of the building

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