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

Answer:

The velocity of the camera is 33.11 m/s.

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

$v^2=(-10.8)^2+2\times(-9.8)\times(-50)$

$v=\sqrt{1096.64}$

$v=33.11\ m/s$

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

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