Question

A projectile is fired horizontally with a speed of $98m{s}^{-1}$ from the top of a hill $490m$ high. Find the magnitude of the vertical velocity with which the projectile hits the ground. (Take $g=9.8m/s^2$)

• A. $98m{s}^{-1}$
• B. $40m{s}^{-1}$
• C. $116m{s}^{-1}$
• D. $196m{s}^{-1}$

Answer 1

The correct option is A $98m{s}^{-1}$

Answer is A.
The projectile is fired from the top $O$ of a hill with speed $u=$ $98m{s}^{-1}$ along the horizontal as shown as $OX$. It reaches the target $P$ at vertical depth $OA$, in the coordinate system as shown, $OA=y=490m$

As, $y=\frac{1}{2}g{t}^2$

$490=\frac{1}{2}\times 9.8t^2$

or $.t=\sqrt{100}=10s$

So, the time taken to reach the ground is $10s$.

The horizontal and vertical components of velocity v of the projectile at point $P$ are
$v_x=u=98m{s}^{-1}$

$v_y=u_y+gt=0+9.8\times 10=98m{s}^{-1}$.

Hence, the velocity wit which the projectile hits the ground is $98m/s$.