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

Answers (1)

@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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