With what minimum speed must you toss a 100 ball straight up to hit the 14 -high roof of the gymnasium if you release the ball 1.4 above the ground?
Solve this problem using energy.

Work is the same as change in energy

You can figure out how much gravitational potential energy the ball has when it is 15.0 m above the ground by using

W = KE = mgh

Difference of the heights;

14-1.4=12.6

W = KE = (.100 kg) * (9.81 m/s²) * (12.60 m) = 12.3606 J

That’s how much kinetic energy the ball will need to reach a height of 14.0 meters

There’s another equation for kinetic energy which is KE = 1/2 m v²

Use that to solve for velocity

v² = KE / (1/2 * m)
v = √ (KE / (1/2 * m) )
= √ (12.606 J / (1/2 * .100kg) )
= 15.87m/s

The speed that it is shot up will also be the speed that it hits the ground, since energy is conserved

Vf² = 274

Vf= √274

Vf= 17 m/s