Question

A bomber is flying horizontally with a speed of 200 m/s. In order to make a direct hit, the bomber drops his bombs 2828 meters in advance of the target. What was the altitude of the plane?

Answer 1

Given data:

* The initial horizontal velocity of the bomber is 200 m/s.

* The horizontal range of the bomb is 2828 m.

Solution:

By the kinematics equation for the horizontal motion of the bomb is,

`R=ut+(1)/(2)at^2`

where u is the initial horizontal velocity, a is the acceleration along the horizontal direction, t is the time taken by the bomb to reach the ground, and R is the horizontal range of the bomb,

Substituting the known values,

`\begin{gathered} 2828=200* t \\ t=(2828)/(200) \\ t=14.14\text{ seconds} \end{gathered}`

Thus, the time taken by the bomb to reach the ground is 14.14 seconds.

By the kinematics equation for the motion along vertical direction is,

`h=u_yt+(1)/(2)gt^2`

where u_y is the vertical component of velocity, t is the time taken to reach the ground, g is the acceleration due to gravity, and h is the height of the bomb,

Substituting the known values,

`\begin{gathered} h=0+(1)/(2)*9.8*(14.14)^2 \\ h=979.7\text{ m} \end{gathered}`

Thus, the height (oof the plane (or bomb before dropping) is 979.7 meters.