Question

A camera of weight 8 newtons is dropped from a drone at a height of 19.9 meters and enters free fall. Assuming no air resistance, what is the final velocity of the camera a moment before it shatters on the ground?

Answer 1

There is no air resistance, the motion of the camera is only due to gravitational force.

Acceleration due to gravity = g = 9.8 m/s²

Weight of camera = 8N

Height from which the camera is dropped = s = 19.9 m

Camera is dropped =>

Initial velocity of the camera = u = 0 m/s

This is the motion of an object under constant acceleration.

The velocity of an object under accelerated motion is given by

v = u + at

The distance traveled by the object under constantly accelerated motion is given by

A = ut + 1/2 at²

where u = initial velocity

a= constant acceleration and

t = time of motion

In this question, distance traveled = height of drone = 19.9

Acceleration = a = g = 9.8 m/s²

u = initial velocity = 0

Putting all the values in the equation of motion

s = ut + 1/2at²

=> 19.9 m = 0*t + 1/2* 9.8 * t²

=> t² = 19.9*2/9.8

=> t² = 4.061

=> t = 2.015

=> This is the time of motion.

Time, before it shatters, is the time of motion.

Velocity of the object is  

v (m/s) = u + at

v (m/s) = 0 + 9.8t

= 9.8*2.015 = 19.75 m/s

Answer: Velocity of the camera before it shatters = 19.75 m/s

Note: I. Have assumed acceleration due to gravity equal to 9.8 m/s².