Question

A 4kg cannonball falls 3m to the ground. How fast will it be moving just before it hits the ground?

Answer 1

The problem can be solved from energy conservation

Let the datum be at ground

Initial potential energy+initial kinetic energy=final potential energy+final kinetic energy

Mass of cannon=m=4 kg

Initial height above ground=h=3 m

Initial velocity=u=0 m/s

Final height above ground=0

Final velocity=v

Acceleration due to gravity=g=9.8 m/s²

Potential energy is given by mass*acceleration due to gravity*height

Kinetic energy is given by 0.5*m*v²

Therefore,

4*9.8*3+0.5*4*0²=4*9.8*0+0.5*4*v²

v=7.67 m/s

Therefore, velocity of cannon ball just before hitting ground is 7.67 m/s