Question

A helicopter is ascending vertically with a speed of 8.0ms^(-1) .At a height of 120m above the earth, a package is dropped from a window.How much time does it take for the package to reach the ground?

1) 1.23s 2) 3.23s 3) 5.83s 4) 7.83s ​

Answer 1

Refer the attachment

Height and Time when package acending:

$h = \frac{ {u}^{2} }{2g } = \frac{8 \times 8}{2 \times 9.8} = 3.26m \\ t(1) = \frac{u}{g} = \frac{8}{9.8} = 0.81s$

Time when package is decending:

$t(2) = \sqrt{ \frac{2(h + 120)}{g} } = \sqrt{ \frac{2(3.26 + 120)}{9.8} } \\ = \sqrt{ \frac{123.26}{4.9} } = \sqrt{25.15} = 5.01s$

total time t = t(1) + t(2)

= 0.81 + 5.01 ≈ 5.83s

Answer 2

Given: A helicopter is ascending vertically with a speed of 8 m/s and the correct height is 12 m instead of 120 m.

To find: How much time does it take for the packet to reach the ground?

Solution:

We have given that A helicopter is ascending vertically with a speed of 8 m/s, so the initial velocity is 8 m/s.

At a height of 12 m above the earth a packet is dropped from a window, so the height is 12 m.

Let the time taken be n (as given in the question)

Lets consider a = -g = -10 m/s² (as coming down so negative)

Now we have the formula:

          s = un + 1/2 x (-a)n²

         -12 = 8n - 1/2 x 10n²

         -12 = 8n - 5n²

          5t² - 8n - 12 = 0

          n = 2.53 seconds