# If a projectile is projected at angle theta of horizontal and it just passed by touching the tip of two walls of height a,seperated by a distance 2a,then show that range of its motion will be 2a cot(theta/2) ?

Mar 16, 2018

Here the situation is shown below,

So,let after time $t$ of its motion,it will reach height $a$,so considering vertical motion,

we can say,

$a = \left(u \sin \theta\right) t - \frac{1}{2} g {t}^{2}$ ($u$ is the projection velocity of projectile)

Solving this we get,

$t = \frac{2 u \sin {\theta}_{-}^{+} \sqrt{4 {u}^{2} {\sin}^{2} \theta - 8 g a}}{2 g}$

So,one value (smaller one) of $t = t$ (let) is suggesting the time to reach $a$ while going up and the other (larger one) $t = t '$ (let) while coming down.

So,we can say in this time interval the projectilw horizontally travelled distance $2 a$,

So,we can write, $2 a = u \cos \theta \left(t ' - t\right)$

Putting the values and arranging,we get,

${u}^{4} {\sin}^{2} 2 \theta - 8 g a {u}^{2} {\cos}^{2} \theta - 4 {a}^{2} {g}^{2} = 0$

Solving for ${u}^{2}$,we get,

${u}^{2} = \frac{8 g a {\cos}^{2} {\theta}_{-}^{+} \sqrt{64 {g}^{2} {a}^{2} {\cos}^{4} \theta + 16 {a}^{2} {g}^{2} {\sin}^{2} 2 \theta}}{2 {\sin}^{2} 2 \theta}$

Putting back $\sin 2 \theta = 2 \sin \theta \cos \theta$ we get,

${u}^{2} = \frac{8 g a {\cos}^{2} {\theta}_{-}^{+} \sqrt{64 {g}^{2} {a}^{2} {\cos}^{4} \theta + 64 {a}^{2} {g}^{2} {\sin}^{2} \theta {\cos}^{2} \theta}}{2 {\sin}^{2} 2 \theta}$

or, ${u}^{2} = \frac{8 g a {\cos}^{2} \theta + \sqrt{64 {g}^{2} {a}^{2} {\cos}^{2} \theta \left({\cos}^{2} \theta + {\sin}^{2} \theta\right)}}{2 {\sin}^{2} 2 \theta} = \frac{8 g a {\cos}^{2} \theta + 8 a g \cos \theta}{2 {\sin}^{2} 2 \theta} = \frac{8 a g \cos \theta \left(\cos \theta + 1\right)}{2 {\sin}^{2} 2 \theta}$

now,formula for range of projectile motion is $R = \frac{{u}^{2} \sin 2 \theta}{g}$

So,multiplying the obtained value of ${u}^{2}$ with $\frac{\sin 2 \theta}{g}$,we get,

$R = \frac{2 a \left(\cos \theta + 1\right)}{\sin} \theta = \frac{2 a \cdot 2 {\cos}^{2} \left(\frac{\theta}{2}\right)}{2 \sin \left(\frac{\theta}{2}\right) \cos \left(\frac{\theta}{2}\right)} = 2 a \cot \left(\frac{\theta}{2}\right)$