Question

two bodies are thrown with the same initial velocity at angles theta and (90-theta) respectively with the horizontal,then their maximum heights are in ratio?

Answer 1

Refer to the above picture for the solution.

Answer 2

Given our two projectiles that are thrown together first projectile is thrown at an angle theta and the second one is thrown at an angle 90 minus theta

The formula for maximum height in a projectile is  $\mathrm{H} 1 / \mathrm{H} 2=\frac{\sin ^{2} \theta}{\sin ^{2}(90-\theta)}=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\tan ^{2} \theta$

$\mathrm{H}=\frac{u^{2} \sin ^{2} \theta}{2 g}$

So now applying the values and finding the ratios of the two projectiles we get  

$\mathrm{H} 1=\frac{u^{2} \sin ^{2} \theta}{2 g} ; \mathrm{H} 2=\frac{u^{2} \sin ^{2}(90-\theta)}{2 g}$