Answer to Question #124823 in Civil and Environmental Engineering for Shubham Dhumal

Question

A helicopter is moving horizontally at 90 km.p.h. at a height of 200 m towards a target on the ground,
which is intended to be shelled, Estimate the distance from the target , where the shell must be released in
order to hit the target.
Also find the velocity with which the shell hits the target and the direction of shell at the time of hitting
the target.

Answer 1

from the newtons equation of motion :

$s=ut + \dfrac{1}{2}at^{2}$

considering vertical projection initial velocity is = 0

$u= 0$

$200=0(t )+ \dfrac{1}{2}(9.8)t^{2}$

$200 \times 2= 0 + 9.8\times t^2$

$t= \sqrt \dfrac{400}{9.8}= 6.389$ seconds

estimated distance = horizontal velocity x time taken to hit the ground

$s= 25 m/s \times 6.389 seconds = 159.725 m$

for vertical motion the initial velocity =0 , assuming acceleration is 9.8 m/s

final velocity velocity with which the shell hits the target and the direction of shell at the time of hitting the target is obtained by the formula

$$v=u +at$$

$v= 0 + 9.8 \times 6.389= 62.61 m/s$

horizontal velocity will be kept constant at 25 m/s

but there will be downward acceleration of the shell to attain a maximum velocity of 62. 61 m/s at the time of hitting the target.