It moves horizontally. It’s angle of projection is
MY SOLUTION
Time of flight is 4sec
Therefore $usin\theta=20$
Also after 2 sec, the vertical velocity is zero
At one second, the vertical velocity is $\frac{u_0}{\sqrt 2}$
$$0=\frac{u_0}{\sqrt 2} - 10$$ $$u_0=10\sqrt 2$$ That’s as far as i could go, what should I do next?
The velocity vector of the stone at a point with time $t\ge0$ after projection is $$(u\cos{(\theta)},u\sin{(\theta)}-gt)$$ When $t=2$, the horizontal component is zero so $$u\sin{(\theta)}-2g=0$$ $$\therefore u=\frac{2g}{\sin{(\theta)}}$$ When $t=1$ both the horizontal and vertical components are equal as they make a $45^\circ$ with the horizontal so $$u\cos{(\theta)}=u\sin{(\theta)}-g$$ $$u\cos{(\theta)}=\left(\frac{2g}{\sin{(\theta)}}\right)\sin{(\theta)}-g$$ $$u\cos{(\theta)}=g$$ $$\therefore u=\frac{g}{\cos{(\theta)}}$$ Combining these equalities we get $$\frac{2g}{\sin{(\theta)}}=\frac{g}{\cos{(\theta)}}$$ $$\sin{(\theta)}=2\cos{(\theta)}$$ $$\tan{(\theta)}=2$$ $$\therefore\theta=\arctan{(2)}$$
