Question

A ball is thrown horizontally from the top of a building 110 m high. The ball strikes the ground 70 m horizontally from the point if release. What is the speed of the ball just before it strikes the ground? Answer in units of m/s.

Answer 1

See below

Step-by-step explanation:

First find the time required to hit the ground :

d = 110 = 1/2 a t^2 a = 9.81 m/s^2

t = 4.74 seconds < ==== use this to find the horizontal velocity

70 m / 4.74 s = 14.8 m/s horizontal

You also need the VERTICAL velocity :

vf = at = 9.81 (4.74) = 46.5 m/s vertical

Now you have the horizontal and vertical components

find the resultant = sqrt ( 14.8 ^2 + 46.5^2) = 48.8 m/s