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Question: A projectile is thrown at angle b with vertical. It reaches a maximum height H. The time taken to re...

A projectile is thrown at angle b with vertical. It reaches a maximum height H. The time taken to reach the highest point of its path is –

A

Hg\sqrt{\frac{H}{g}}

B

2Hg\sqrt{\frac{2H}{g}}

C

H2g\sqrt{\frac{H}{2g}}

D

2Hgcos⁡β\sqrt{\frac{2H}{g\cos\beta}}

Answer

2Hg\sqrt{\frac{2H}{g}}

Explanation

Solution

H = u2cos⁡2β2g\frac { u ^ { 2 } \cos ^ { 2 } \beta } { 2 g } ̃ 4cosb = 2gH\sqrt { 2 \mathrm { gH } }

t = ucos⁡βg\frac { u \cos \beta } { g } = 2gHg\frac { \sqrt { 2 \mathrm { gH } } } { \mathrm { g } } ̃ t = 2Hg\sqrt { \frac { 2 \mathrm { H } } { \mathrm { g } } }