Question

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically upward from ground level, such that it has zero speed when it reaches the top of the building.

a) When do the two balls pass each other? Answer it in terms of h.

b)Which ball has greater speed when they are passing?

c)What is the height of the two balls when they are passing?

Answer 1

? At the point where the two balls pass each other, their respective velocities will be equal in magnitude but opposite in direction.
? This point would be at a height of 3H/4 from the ground.

Solution:
Let the initial velocity of the 2nd ball be U.
Then, its maximum height H = (U^2) / (2g) [where g denotes acceleration due to gravity]
This maximum height is the same point where its velocity will be zero, i.e., top of the building

If we observe this casse of motion through the frame of reference of the 1st ball, then the 1st ball would appear to be in a state of rest and the 2nd ball would appear to move towards the first ball in a straight line with velocity U and net acceleration zero.
? The 2nd ball will cover a distance H with a constant speed of U.
? The time at which the two balls will pass each other = H / U
By substituting the value of H, we get time as [ U / (2g) ]

Let:
V1 be the velocity and S1 be the distance from ground for the 1st ball
V2 be the velocity and S2 be the distance from ground for the 1st ball

Then,
V1 = U - gt ..........................[Eq. 1]
S1 = Ut - g*t*t/2 ...................[Eq. 2]

V2 = -gt ..............................[Eq. 3]
S2 = [ (U^2) / 2g ] - g*t*t/2 ....[Eq. 4]

By substituting the value of time in Eq.1 and Eq.3 , we get
V1 = U/2
V2 = -U/2 [ minus sign indicates downward direction ]

By substituting the value of time in Eq.2 and Eq.4 , we get
S1 = S2 = 3H/4