A 5 kilograms bowling ball is dropped out a window. It hits the ground, and bounces upward. The velocity change of the ball is noted to be 15 m/s downward and 12 m/s upward. What is the contact time for the ball if the force applied on the ball from the ground is equal to 10 N?

0.11 m/s and 0.28 m/s should be correct

massxgravity x heght=weightxheight= pe

4900=490xh

h=10

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer