Question

A parachutist with a camera, both descending at a speed of 10.8 m/s, releases that camera at an altitude of 50 m. In this problem, take "up" to be positive.

Answer 1

The velocity of the camera is 33.11 m/s.

Step-by-step explanation:

Given that,

Speed = 10.8 m/s

Altitude = 50 m

Suppose determine the velocity of the camera just before it hits the ground?

We need to calculate the velocity of the camera

Using equation of motion

`v^2=u^2+2gs`

Where, v = final velocity of camera

u = initial speed of camera

s = distance

Put the value into the formula

`v^2=(-10.8)^2+2*(-9.8)*(-50)`

`v=√(1096.64)`

`v=33.11\ m/s`

The direction will be downward so it is the negative velocity.

Hence, The velocity of the camera is 33.11 m/s.