Question

A helicopter is ascending vertically at a steady speed of 15 ms-1 when a package is dropped from it at
a height of 90 m above the ground. Calculate
(i) The speed of impact of the package at the ground.
(ii) The time taken for the package to fall to the ground.[g=9.8ms-2
]

Answer 1

Given:

A helicopter is ascending vertically at a steady speed of 15 m/s when a package is released.

To find:

• Speed of impact on ground
• Time taken to reach ground

Calculation:

We must understand one important point :

• When the package is released , its initial velocity will be 15 m/s upwards, so it's sign will be opposite to the sign of gravity (i.e. g).

Let velocity of impact be v :

Applying EQUATIONS OF KINEMATICS:

$\sf \: {v}^{2} = {u}^{2} + 2ah$

$\sf \implies\: {v}^{2} = {( - 15)}^{2} + 2g( 90)$

$\sf \implies\: {v}^{2} = 225+ 2 \times 10 \times 90$

$\sf \implies\: {v}^{2} = 225+ 1800$

$\sf \implies\: {v}^{2} = 2025$

$\sf \implies\: v = 45 \: m {s}^{ - 1}$

So, velocity of striking ground is 45 m/s.

• Now, let time taken be t :

$\sf \: v = u + at$

$\sf \implies \: 45 = - 15 + (10 \times t)$

$\sf \implies \: 10t = 60$

$\sf \implies \: t = 6 \: sec$

So, time taken to reach ground is 6 seconds.