Question

A ball is thrown from ground level so as to just clear a wall $4$ metres high at a distance of $4m$ and falls at a distance of $14m$. From the wall, then the magnitude of the velocity of the ball is-

• A. $\sqrt{281}$ m/sec
• B. $\sqrt{812}$ m/sec
• C. $\sqrt{182}$ m/sec
• D. None of the above

Answer 1

The correct option is C $\sqrt{182}$ m/sec

The ball passes through the point $P(4,4)$. So it's range$=4+14=18m.$

The trajectory of the ball is,

$y=x\times tan\theta (1-x/R)$

Now $x=4m,y=4m,andR=18m$

$\therefore 4=4\times tan\theta [1-4/18]=4\times tan\theta \times 7/9⟹tan\theta =9/7$

And $R=\frac{2\times u^2\times sin\theta \times cos\theta }{g}$

$⟹18=\frac{2}{9.8}\times u^2\times \frac{9}{\sqrt{130}}\times \frac{7}{\sqrt{130}}$

$⟹u=\sqrt{182}$ $m/sec$