Question

A projectile is projected from point O with the velocity $10m/s$ at an angle $60$. A is a point in its trajectory such that OA makes $45$ with horizontal. Then, find OA.

Answer 1

$v^1\cot 45°=u\cos 60°v^1=\frac{u\cos 60°}{\cos 45°}=u\times \frac{1}{2}\times v_2{v}^1=\frac{u}{\sqrt{2}}$

time taken from $0$ to $A$ is $t$

$v^1\sin 45°-u\sin 60°=9t\frac{u}{\sqrt{2}}\times \frac{1}{\sqrt{2}}-\frac{\sqrt{3}}{2}u=-10tt=\frac{\sqrt{3}-1}{20}x=u\cos 60°t=\frac{u}{2}\times \sqrt{3}-120u=\frac{(\sqrt{3}-1)}{40}{u}^2({v^1\sin 45°)}^2-{\left(u\sin 60°\right)}^2=-2gy\frac{u^2}{4}-\frac{3u^2}{4}=-2\times 10yy=\frac{u^2}{40}$

Now

$OA=\sqrt{x^2+y^2}=\sqrt{\frac{u^4}{{40}^2}+\frac{{(\sqrt{3}-1)}^2{u}^2}{{40}^2}}OA=\frac{u^2}{40}\left(\sqrt{3-2\sqrt{3}+1+1}\right)=\frac{u^2}{40}\left(\sqrt{1}25-2\sqrt{3}\right)$

$OA=20$