Question

A 5.0 kg cannonball is dropped from the top of a tower. It falls for 1.6 seconds before slamming into a sand pile at the base of the castle. What is the velocity of the cannonball just before it hits the ground ?

Answer 1

15.7 m/s

Step-by-step explanation:

The motion of the cannonball is a accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground (gravitational acceleration). Therefore, the velocity of the ball at time t is given by:

`v(t)=u + gt`

where

u = 0 is the initial velocity

g = 9.8 m/s^2 is the acceleration

t is the time

If we substitute t=1.6 s into the equation, we find the final velocity of the cannonball:

`v(1.6 s)=0+(9.8 m/s^2)(1.6 s)=15.7 m/s`