Question

Maximum time of flight, for a body projected from
ground, will be attained if angle of projection with
horizontal is

Answer 1

General Equation of time in a projectile is:

$T = \dfrac{2u \sin( \theta) }{g}$

Now, differentiation w.r.t $\theta$:

$\implies \dfrac{dT}{d \theta} = \dfrac{2u}{g} \times \dfrac{d \{\sin( \theta) \}}{d \theta}$

$\implies \dfrac{dT}{d \theta} = \dfrac{2u \cos( \theta) }{g}$

• For Maxima, 1st order derivative is equal to zero.

$\implies \dfrac{dT}{d \theta} = \dfrac{2u \cos( \theta) }{g} = 0$

$\implies \cos( \theta) = 0$

$\implies \: \theta = {90}^{ \circ}$

Now, 2nd order derivative:

$\implies \dfrac{ {d}^{2} T}{d{ \theta}^{2} } = \dfrac{2u}{g} \times \dfrac{d \{ \cos( \theta) \} }{d \theta}$

$\implies \dfrac{ {d}^{2} T}{d{ \theta}^{2} } = - \dfrac{2u \sin( \theta) }{g}$

$\implies \dfrac{ {d}^{2} T}{d{ \theta}^{2} } \bigg| _{ \theta = {90}^{ \circ} } = - \dfrac{2u }{g}$

$\implies \dfrac{ {d}^{2} T}{d{ \theta}^{2} } \bigg| _{ \theta = {90}^{ \circ} } < 0$

So, maximum time is for angle of projection = 90°.

Answer 2

89 degree because of the question

ok thanks