Question

A car is moving horizontally with velocity $v$. A shell is fired upward with velocity $u$ inclined at angle $\theta$ with the horizontal. The horizontal range of the shell related to ground is

• A. $\frac{2u^{2}v}{g}$
• B. $\frac{2v^{2}u}{g}$
• C. $\frac{2u^2{v}^2}{g}$
• D. $\frac{2(v+u\cos \theta )u\sin \theta }{g}$

Answer 1

The correct option is D $\frac{2(v+u\cos \theta )u\sin \theta }{g}$

A shell is fired upward with velocity u , inclined at angle $\theta$

with the horizontal

The horizontal component for shell will be

$=u\cos \theta$

Velocity of car is give as $\nu$

Hence total horizontal velocity component, $v+u\cos \theta ⟶1$

Let the time taken by the projectile motion be $t$, when the shells reaches ground, the vertical displacement is zero. Also from second equation of motion

We know that,

$y=ut\sin \theta -\frac{1}{2}g{t}^2$

We are considering vertical component of velocity of the shell, putting $y=0$ in second equation of motion we get

$t=\frac{2u\sin \theta }{g}⟶2$

Range is horizontal distance traveled which will be the product of horizontal component of velocity $s$ time

$\therefore$ Range$=(v+u\cos \theta )t$

Substituting values of $t$ from equation $2$

Range$=\frac{2(v+u\cos \theta )u\sin \theta }{g}$