Question

The height of a helicopter above the ground is given by h = 3.15t3, where h is in meters and t is in seconds. At t = 2.10 s, the helicopter releases a small mailbag.
How long after its release does the mailbag reach the ground?

Answer 1

Answer:

The mailbag will take 2.44 seconds to reach the ground.

Explanation:

The height of a helicopter above the ground is given by:

$h = 3.15\times t^3$

Height of helicopter at t = 2.10 seconds

$h(2.10 )=3.15\times (2.10 )^3 m=29.17 m$

The helicopter releases a small mailbag from the height of 29.17 m.

The initial velocity of mailbag = u = 0 m/s

Duration in which mailbag will reach the ground = T

Acceleration due to gravity = g = $9.8 m/s^2$

Using second equation of motion ;

$s=ut+\frac{1}{2}gt^2$

We have , s = 29.17

u = 0 m/s

t = T

$29.17m=0 m/s\times T+\frac{9.8 m/s^2\times T^2}{2}$

Solving for T, we gte :

T = 2.44 seconds

The mailbag will take 2.44 seconds to reach the ground.