First slide

Projection Under uniform Acceleration

Question

Two balls are projected at an angle $θ and (90^{0} - θ)$ to the horizontal with the same speed. The ratio of their maximum vertical heights is

Moderate

A

1 : 1
B

$\tan θ : 1$
C

1 : $\tan θ$
D

$\tan^{2} θ : 1$

Solution

$H_{1} \frac{= u^{2} \sin^{2} θ}{2 g}$ ---------------(i)
$H_{2} = \frac{u^{2} \sin^{2} (90^{0} - θ)}{2 g} = \frac{u^{2} \cos^{2} θ}{2 g} - - - - - (ii)$
Dividing (i) by (ii), we get $\frac{H_{1}}{H_{2}} = \frac{\tan^{2} θ}{1}$

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