Question

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

• A. 1 : 1
• B. $\tan\theta:1$
• C. $1:\tan\theta$
• D. $\tan^{2}\theta:1$

Answer 1

Answer: (D)

$\tan^{2}\theta:1$

Answer 2

$H_{1} = \frac{u^{2}\sin^{2}\theta}{2g}$ ….. (i)

$H_{2} = \frac{u^{2}\sin^{2}(90{^\circ} - \theta)}{2g} = \frac{u^{2}\cos^{2}\theta}{2g}$ ….. (ii)

Divide (i) by (ii), we get

$\frac{H_{1}}{H_{2}} = \frac{\tan^{2}\theta}{1}$