Question

two tall buildings are situated 200m apart, with what speed must a ball be thrown horizontally from the window 540m above the ground in one building, so that it will enter a window 50m above the ground in the other?

Answer 1

Vertical distance between the two buildings = 540 – 50 = 490m

Horizontal distance = 200 m

Suppose ball speed is u and time taken to reach the point is t so

Ball has no initial velocity and takes time t for 490 meter so

S = ut + ½ at^2

490 = 0 + ½ (9.8)(t^2)

T = 10 sec

Horizontal component remain same so we use x = ut

200 = u ( 10 )

U = 10

Ball will be thrown at a speed of 10 m / s

Answer 2

If you wand to findout ball speed let first calculate the distance between buildings.
vertically distance will be= 540- 50= 490 meter.
Horizontal distance will be= 200 meter.
we know that
S= ut +1/2 at^2
here u is the speed of the ball "t" is the time taken to reach
Since there is zero initial velocity of ball "S' is the distance so
490 = 0+1/2(9.8)(t^2)

so upon calculating time  t= 10seconds.
for horizontal components
200 =u (10) if we take x=ut
so u= 10
hence ball speed will be 10m/s