Question

A projectile is fired horizontally with a speed of $98m{s}^{-1}$ from the top of a hill $490m$ high. Find the distance of the target from the hill.

• A. $980m$
  
• B. $490m$
  
• C. $1960m$
  
• D. $840m$
  

Answer 1

The correct option is A $980m$

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, $OA=y=490m$

Since initial vertical velocity is zero.

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

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

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

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

Distance of the target from the hill is given by,

$AP=x=Horizontalvelocity\times time=98\times 10=980m$

Hence, the distance of the target from the hill is $980m$.