Question

A ball projected from ground at an angle of `45^(@)` just clears a wall infront. 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

 Correct option is (B) 2.4 m

Here,

Range = 4 m + 6 m = 10 m

As the ball is projected at an angle 45^∘ to the horizontal,

Therefore, Range = 4H

10 = 4H 

⇒ H = \(\frac {10} 4\) = 2.5 m

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

\(\therefore u^2 = \frac{H \times 2g}{\sin^2\theta}\)

\(= \frac{2.5\times2 \times 10}{\left(\frac 1 {\sqrt 2}\right)^2}\)

\(= 100\)

\(u = \sqrt{100} = 10 ms^{-1}\)

From trajectory formula,

\(y = x \tan\theta - \frac{gx^2}{2u^2\cos^2\theta}\)

So, the height of the wall PA is,

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

\(h = (4 \times 1) - \frac12 \times \frac{10 \times 4^2}{10^2 \times (1/ \sqrt 2)^2}\)

\(∴ h = 2.4 m\)