Question

Prove that for a projectile motion time of flight is twice the time for the maximum height

Answer 1

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Answer 2

Answer:

      To prove that time of flight depends on twice the time of maximum height, we will first see about the time of flight as it is dependent on the Magnitude of initial velocity and angle In which it makes the projectile $T=2\ \times \frac{uy}{g}.$

Now writing main component of you we get

                      $T=2u \times \frac{sin\theta}{g}.$

     And we know the maximum height is the maximum altitude it reaches and is given by $\frac{2 u^2}{2g}$

By equalling, we get

                           $\frac{2u sin\theta}{g} = \frac{2 u^2}{2g}$

                         $2u sin\theta = u^2$

                           $2 sin\theta = u$

     Now we see that doubling this value of maximum height, we get the value of time of flight hence we can say that my flight is twice the time of maximum.