Question

The speed of a projectile when it reaches its maximum height is 0.38 times its speed when it is at half its maximum height.What is the initial projection angle of the projectile?

Answer 1

$mv_1^2/2+mgh/2=mgh+mv_2^2/2$

$v_1^2+gh=2gh+v_2^2$

$v_2^2/0.38^2+gv_0^2\sin^2\alpha/(2g)=2gv_0^2\sin^2\alpha/(2g)+v_2^2$

$v_0^2\cos^2\alpha/0.38^2+gv_0^2\sin^2\alpha/(2g)=2gv_0^2\sin^2\alpha/(2g)+v_0^2\cos^2\alpha$

$\cos^2\alpha/0.38^2+\sin^2\alpha/2=\sin^2\alpha+\cos^2\alpha$

$\cos^2\alpha(1/0.38^2-1)=\sin^2\alpha-\sin^2\alpha/2$

$5.925=0.5\tan^2\alpha\to\tan\alpha=3.442\to\alpha=73.8°$ . Answer