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  • A projectile is thrown at an angle  with vertical .It reaches a maximum height H . The time taken to reach

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 

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

\sqrt{\frac{2H}{g\cos \beta }}


Answers (1)

best_answer

As we have learned

Time of Flight -

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

t=\frac{2u\sin \theta }{g}

here,   \theta = (90-\beta ) \ \ \ \ \ (\text{Angle of projection is taken from horizontal direction})

and

  H = \frac{u^2\sin ^2 \theta }{2g} =\frac{u^2\sin ^2(90-\beta) }{2g}= \frac{u^2\cos ^2 \beta }{2g}\; \; \\ \\ or \; \; (u\cos \beta )^2= 2gH

i.e u\cos \beta = \sqrt{2gH}........(1)

Time is taken to reach the highest point

\\ T=\frac{u\sin \theta }{g}=\frac{usin(90-\beta) }{g}= \frac{u\cos \beta}{g} = \sqrt{ \frac{2H}{g}}

 

Posted by

shivangi.shekhar

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