Question

A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away from the wall, the angle of projection of ball is

• A. tan^-1 3/2
• B. tan^-1 2/3
• C. tan^-1 ½
• D. tan^-1 ¾

Answer 1

Answer: (B)

tan^-1 2/3

Answer 2

Using R = u²sin 2θ/g we get

g/u² = sin 2θ/R

Range, R = 6 + 18 = 24 m

$\therefore$ g/u² = sin 2θ/24

As y = x tan θ - (gx²/2u² cos² θ)

for x = 6m, y = 3 m

and g/u² = sin 2θ/24 we get

3 = 6 tanθ - $\frac{\sin 2\theta}{2 \times 24}\frac{6^{2}}{\cos^{2}\theta}$

Using sin 2θ = 2 sin θ cos θ, we get

tan θ = 2/3 or θ = tan^-1 (2/3)