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A projectile is thrown at an angle beta with vertical

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

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@Harshit

As we have learned

Time of Flight -

Time for which projectile remains in the air above the horizontal plane.

- wherein

$t=\frac{2U\sin \Theta }{g}$

SO

Time taken to reach highest point =$\frac{t}{2}=\frac{U\sin \beta }{g}$

$H = \frac{u^2\sin ^2 \beta }{2g} \; \; or \; \; (u\sin \beta )^2= 2gH$

$u\sin \beta = \sqrt{2gH}........(1)$

Time taken to reach highest point = $u\sin \beta/ g = \sqrt{2H/g}$

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