Question

A ball is thrown horizontally from the top of

a building 130 m high. The ball strikes the

ground 64 m horizontally from the point of

release.

What is the speed of the ball just before it

strikes the ground?

Answer in units of m/s.

Answer 1

Answer: 50.4 m/s

Explanation:

height of the building = 130 m

horizontal distance = 64 m

acceleration due to gravity = 9.8 m/s

applying the equation of motion s = ut + (0.5at^2) and assuming that air resistance is neglected, we first find the time it takes to hit the ground

where s = distance (height of the building)

u = initial velocity (which is 0 in this case )

t = time

a= acceleration due to gravity

 

130 = ( 0 x t) + ( 0.5 x 9.8 x t^2)

130 = 4.9 x t^2

t = $\sqrt{(130 / 4.9)}  $

t = 5.15 secs

now we use the equation Vy = Uy + at to get its speed before hitting the ground

where Uy = initial velocity(0 in this case)

Vy = 0 + (9.8 x 5.15)

Vy = 50.4 m/s