SATHEE: Kinematics Question 511

Kinematics Question 511

Question: A stone projected with a velocity $u$ at an angle $θ$ With the horizontal reaches maximum height $H_{1}$ When it’s projected with velocity u at an angle $(\frac{π}{2} - θ)$ with the horizontal, it reaches maximum height $H_{2}$ . The relation between the horizontal range R of the projectile, $H_{1}$ and $H_{2}$ it’s

Options:

A) $R = 4 \sqrt{H_{1} H_{2}}$

B) $R = 4 (H_{1} - H_{2})$

C) $R = 4 (H_{1} + H_{2})$

D) $R = \frac{H_{1}^{2}}{H_{2}^{2}}$

Show Answer

Answer:

Correct Answer: A

Solution:

[a] $H_{1} = \frac{u^{2} \sin^{2} θ}{2 g} a n d H_{2} = \frac{u^{2} \sin^{2} (90 - θ)}{2 g} = \frac{u^{2} \cos^{2} θ}{2 g}$

$H_{1} H_{2} = \frac{u^{2} \sin^{2} θ}{2 g} \times \frac{u^{2} \cos^{2} θ}{2 g}$

$= \frac{{(u^{2} s i n 2 θ)}^{2}}{16 g^{2}} = \frac{R^{2}}{16}$

$[∴ R = 4 \sqrt{H_{1} H_{2}}]$

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