A ball is thrown horizontally from the roof of a building 45.0 m tall and lands 24.0 m from the base.

what was the ball's initial speed?

when u throw a ball horizontaly from a 45 m tall building ur horizontal velocity do not change becoz gravity is downwards and speed is horizontal .since horizontal velocity is not changing so we can apply this formula distance=speed*time
in this case it is horizontal velocity*time=distance here distance is 24m (horizontal distance)

now we do not have the value of time.

but we can find out
as i said gravity is downwards so when u throw the ball it starts accelerating downwards and ball gains some vertical velocity .note intial vertical velocity was zero becoz we have given horizontal velocity to ball.
so for vertical velocity we can apply

h=ut +1/2at^2
u=0
t=(2h/g)^(1/2) here h is height of building .(vertical distance)
now u have got the time
plug the value in 1st eqn u will get the ans

since the throw was horizontal, there is no vertical component of the initial velocity.
this means that time for ball to fall to ground is same as if it was just released (instead of being thrown horizontally).

y=0.5*at^2

45=0.5*9.8*t^2
t=3.03s

there is no force in horizontal direction (assuming no wind and no air resistance) so horizontal velocity remains constant. travelled distance in horizontal direction was
x=v*t
24=v*3.03
v=24/3.03=7.92 m/s

8 m/s