Question

An aeroplane is flying horizontally at height of 490 m above a point P on ground and a packet is to be dropped from it, which should reach at a point Q on ground which is 1600 m away from point P. What should be the speed of plane at the time of release of packet?
(g = 9.8 m s–2)

Answer 1

160 m/s

Explanation:

Assume gravitational acceleration to be 9.8 m/s^2.

Since the aeroplane is 490m in the sky, let t be the time it takes for the packet to reach the ground.

$0.5*g*t^{2} = 490$

And

$g = 9.8m/s^2$

This gives

$t^{2} =$ 100s

$t = 10 seconds$

Thus, the horizontal speed of the plane should be such that it covers 1600meters in 10 seconds.

Speed = distance/time

This gives

Vh = 1600/10 metres per second

Therefore our answer is 160m/s.

Answer 2

Given:

Height = 490m

Distance of point Q = 1600,

g = 9.8 ms–2

To find:

Speed of plane at the time of the release of packet

Solution:

Calculating, the time taken by the bomb to fall through a height -

t = √2h/g

Substituting the value -

t = √ 2 × 490/9.8

t = √980/9.8

t = 10s

Now,

The horizontal speed of the plane should be such that it covers 1600meters in 10 seconds.

Using the distance formula -

Speed = distance/time

  = 1600/10

= 160

Answer: Speed of plane at the time of the release of packet is 160m/s.