A ball is thrown from the ground to clear a wall 3m high at a...

We know that range of a projectile

$R =u \dfrac{2\sin 2\theta}{g}$

Here $R = 6 + 18 = 24$

$∴ \dfrac{u^{2}\sin 2\theta}{g} = 24$ ...... (i)



The equation trajectory of a projectile

$y = xtan \theta - \dfrac{gx^{2}}{2u2\cos^{2} \theta}$

$3 = 6tan \theta - \dfrac{36g}{2u2\cos^{2} \theta}$ ...... (ii)

From equation (i), $\dfrac{g}{u^{2}} = \dfrac{\sin 2\theta}{24}$

$= \dfrac{sin \theta cos \theta}{12}$

Substituting in equation (ii), we get

$3 = 6tan \theta - \dfrac{3}{2}tan \theta$

$3 = \dfrac{9}{2}tan \theta$

$⇒$\theta = \tan^{- 1} \left(\dfrac{2}{3}\right)