Question

A ball projected from the ground at an angle of ${45}^{\circ }$just clears a wall in front. If the point of projection is $4\text{m}$ from the foot of the wall and the ball strikes the ground at a distance of $6\text{m}$ on the other side of the wall, the height of the wall is :

• A. $4.4\text{m}$
• B. $2.4\text{m}$
• C. $3.6\text{m}$
• D. $1.6\text{m}$

Answer 1

The correct option is B $2.4\text{m}$

Here,
$\text{Range}=4\text{m}+6\text{m}=10\text{m}$

As the ball is projected at an angle ${45}^{\circ }$ to the horizontal,

Therefore, $\text{Range}=4H$

$10=4H\Rightarrow H=\frac{10}{4}=2.5\text{m}$

Maximum height, $H=\frac{u^2{\sin }^2\theta }{2g}$

$\therefore u^2=\frac{H\times 2g}{{\sin }^2\theta }=\frac{2.5\times 2\times 10}{{\left(\frac{1}{\sqrt{2}}\right)}^2}=100$

$u=\sqrt{100}=10{\text{ms}}^{-1}$

From trajectory formula,

$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

So, the height of the wall $\text{PA}$ is,

$h=\left(OA\right)\tan \theta -\frac{g(OA{)}^2}{2u^2{\cos }^2\theta }$

$h=(4\times 1)-\frac{1}{2}\times \frac{10\times 4^2}{{10}^2\times (1/\sqrt{2}{)}^2}$

$\therefore h=2.4\text{m}$

Hence, option $\left(\text{B}\right)$ is correct.