Question

A helicopter is ascending vertically with a speed of 5.40m/s at a height of 105m above the earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground? (Hint: What is Vo for the package?)

Answer 1

Given:

u = - 5.40m/s

s = 105m

To find:

t =?

Explanation:

A helicopter is ascending vertically with a speed of 5.40m/s at a height of 105m are on top of the earth, a package is dropped from the helicopter.

Hence acceleration due to gravity is positive due to vertically upward direction.

i.e. g = 9.81

By the first law of kinematics

s = ut + 0.5a$t^2$

105 = ( - 5.40 ) ( t ) + 0.5 × 9.81 × t^2

4.90 $t^2$- 5.40 t - 105 = 0

solving above eq, we get,

t = 5.2128     t = -4.1107

But t ≠  -4.1107

∴ t =  5.2128 s

5.2128 s are required to take for the package to reach the ground.