Question

A ball is thrown horizontally from a height of 17.7 m and hits the ground with a speed that is 3.00 times its initial speed. What is the speed in the vertical direction just before the ball hits the ground

Answer 1

19.76 m/s

Step-by-step explanation:

Let v be the speed just before the ball hits the ground. The v/3 is the initial speed. When the ball falls to the ground its potential energy is converted to kinetic energy:
`E_p + E_(k0) = E_k`
`mgh + mv_0^2/2 = mv^2/2`where m is the mass and h = 17.7 is the vertical distance traveled,
`v_0 = v/3` is the initial velocity , and g = 9.81 m/s2 is the gravitational acceleration. We can divide both sides by m:

`gh + v^2/18 = v^2/2`

`gh = 9v^2/18 – v^2/18 = 8v^2/18 = 4v^2/9`

`9.81*17.7 = 4v^2/9`

`v^2 = 390.68`

`v = √(390.68) = 19.76m/s`