Question

Two Buildings A and B are each of height H as measured from the ground and are located a distance L apart as shown in the figure. A ball is shot horizontally with a velocity v0 ​from building A such that it just hits the middle of building B before hitting the ground. At what new velocity v' would the ball have to be shot from building A such that the ball now hits Building B one-quarter of the way up from the bottom:

Answer 1

For the first case,
The initial horizontal velocity is v0
The initial vertical velocity is 0.
The vertical distance travelled is d = H/2
The horizontal distance travelled = L
We can equate the time taken for the vertical and horizontal flights.
The time taken to cover the vertical distance,
from the equation, d = ut + (1/2) at2
t = sqrt (2 d / g) = sqrt (H/g) ------------------ 1
The time taken to cover the horizontal distance L is,
t = L / v0 -------------------------- 2
Equating 1 and 2
sqrt (H/g) = L / v0
H = g (L / v0)2
For the second case,
The ball hits 3/4 distance from the top of the building, using the above equations
sqrt (3H/4g) = L / v'
v' = L / sqrt (3 H /4 g)
Substituting for H from the previous equation, we get
v' = L / sqrt ( 3g (L / v0)2 / 4g)
v' = sqrt (4/3) v0