Question

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

• A. $4.4m$
• B. $2.4m$
• C. $3.6m$
• D. $1.6m$

Answer 1

Answer: (B)

As ball is projected at an angle $45^{°}$ to the horizontal therefore Range $=4H$
or $10=4H\Rightarrow H=\frac{10}{4}=2.5m$
$\left(\because Range=4m+6m=10m\right)$
Maximum height, $H=\frac{u^{2}si{n}^2\theta }{2g}$
$\therefore u^2=\frac{H\times 2g}{si{n}^2\theta }=\frac{2.5\times 2\times 10}{{\left(\frac{1}{\sqrt{2}}\right)}^2}=100$
or, $u=\sqrt{100}=10m{s}^{-1}$
Height of wall $PA$
$=OAtan\theta -\frac{1}{2}\frac{g{\left(OA\right)}^2}{u^{2}co{s}^2\theta }$
$=4-\frac{1}{2}\times \frac{10\times 16}{10\times 10\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}}=2.4m$