Question

A 27.5 kg ball is thrown at an angle of 22.7° above the horizontal off of a building such that its initial potential energy is 423 J, and the initial kinetic energy is 347 J.

What will be the speed of the ball at the moment before it hits the ground?

*Remember to use the conservation of Energy!

Answer 1

7.48 m/s

Step-by-step explanation:

Total energy of the ball doesn't change

So,

Initial energy = Final energy

Initial Kinetic + Initial Potential = Final Kinetic + Final Potential

347 + 423 = ( (1/2) * 27.5 * v^2 ) + ( 27.5 * 9.81 * 0 )

770 = 13.75 * v^2

v^2 = 56

v = 7.48