Question

A projectile is projected with a velocity V0 making an angle θ with horizontal. Derive equations for maximum height and time of flight for a projectile.​

Answer 1

Time and Max Height of Projectile:

• Let body is projected at angle θ with horizontal with velocity v_(0).

Along Y axis :

$y = u_{y}(t) - \dfrac{1}{2} g {t}^{2}$

$\implies \: y = v_{0} \sin( \theta) t - \dfrac{1}{2} g {t}^{2}$

For total time , the displacement in Y axis should be zero.

$\implies \: 0 = v_{0} \sin( \theta) t - \dfrac{1}{2} g {t}^{2}$

$\implies \: v_{0} \sin( \theta) t = \dfrac{1}{2} g {t}^{2}$

$\boxed{ \implies \: t = \dfrac{2v_{0} \sin( \theta) }{g} }$

Now, for max height :

• We will put value of t/2 as max height is reached at half time.

$\implies \: y_{max} = v_{0} \sin( \theta) \dfrac{t}{2} - \dfrac{1}{2} g { (\dfrac{t}{2} )}^{2}$

$\implies \: y_{max} = \dfrac{ {v_{0}}^{2} { \sin}^{2} ( \theta)}{g} - \dfrac{ {v_{0}}^{2} { \sin}^{2} ( \theta)}{2g}$

$\boxed{ \implies \: y_{max} = \dfrac{ {v_{0}}^{2} { \sin}^{2} ( \theta)}{2g} }$

Hope It Helps.