Question

A ball is thrown from ground level so as to just clear a wall $4\text{m}$ high at a distance of $4\text{m}$ and falls at a distance of $14\text{m}$ from the wall. Find the magnitude of the initial velocity (in $\text{m/s}$) of the ball.

[Upto two decimal places]

Answer 1

The ball passes through the point $\text{P}(4,4)$.

So from figure, its range $R=4+14=18\text{m}$.


The equation of projectile is,

$y=x\tan \theta [1-\frac{x}{R}]$

Now, $x=4\text{m},y=4\text{m}$ and $R=18\text{m}$

$\therefore 4=4\tan \theta [1-\frac{4}{18}]$

$\Rightarrow 4=4\tan \theta \times \frac{7}{9}$

$\Rightarrow \tan \theta =\frac{9}{7}$

$\Rightarrow \sin \theta =\frac{9}{\sqrt{9^2+7^2}}=\frac{9}{\sqrt{130}}$ and

$\cos \theta =\frac{7}{\sqrt{130}}$

As, we know that

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

$\Rightarrow u^2=\frac{Rg}{\sin 2\theta }=\frac{Rg}{2\sin \theta \cos \theta }$

Substituting the values,

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

$\Rightarrow u^2=182$

$\therefore u=\sqrt{182}=13.49\text{m/sec}$

Accepted answers :$13.49,13.50$