An observer having a gun observes a remotely controlled balloon. When he first noticed the balloon, it was at an altitude of `800m` and moving vertically upward at a constant velocity of `5m//s`. The horizontal displacement of balloon from the observer is `1600m`. Shells fired from the gun have an initial velocity of `400 m//s` at a fixed angle `theta ( sin theta =3//5 `and `cos theta =4//5 )`. The observer having gun waits `(` for some time after observing the balloon `)` and fires so as to destroy the balloon. Assume `g=10m//s^(2)`. Neglect air resistance.

The altitude of the collosion above ground level is

Options

A. `1250m`
B. `1325m`
C. `1075m`
D. `1200m`

Answer

C

Explanation

`(`Easy `)` The motion in the `x-` direction is a constant velocity motion. We find the flight time `=(1600m)/(u_(x))`
`(1600)/(400costheta)=5 sec.`
Flight time `=5 sec.`
`(` Easy `)` From the flight time, the initial velocity in the `y-` direction and the acceleration in the `y-` direction, we can calculate the altitude of the shell`:`
`h=u_(x)t-(1)/(2)g t ^(2)=(1200)/(5)xx5-(1)/(2)xx10xx25=1200-125=1075m`
height `=1075m`