A parachutist with a camera, both descending at a speed of 10.8 m/s, releases that camera at an altitude of 60 m. In this problem, take "up" to be positive. What is the velocity of the camera just before it hits the ground?

The velocity of the camera just before it hit the ground = 35.97 m/s.

Explanation:

Velocity: This can be defined as the ratio of the displacement of a body to the time. Velocity is a vector quantity, and as such it can be represented both in magnitude and direction.

From the equation of motion,

v² = u² + 2gs ... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, s = distance.

Note: Before the velocity of the camera before it hits the ground = The final velocity of the camera.

Given: u = 10.8 m/s, s = 60 m. g = 9.81 m/s.

Substituting into equation 1,

v² = 10.8² + 2 (9.81) (60)

v² = 116.64+1177.2

v² = 1293.84

v = √ (1293.84)

v = 35.97 m/s.

Hence the velocity of the camera just before it hit the ground = 35.97 m/s.