The-speed-of-a-projectile-when-it-is-at-its-greatest-height-is-2-5-times-its-speed-at-half-the-maximum-height-What-is-its-angle-of-projection-

Question Number 16455 by Tinkutara last updated on 22/Jun/17

The speed of a projectile when it is at

its greatest height is \sqrt{\frac{2}{5}} times its

speed at half the maximum height .

What is its angle of projection ?

Answered by sandy_suhendra last updated on 23/Jun/17

A = point of the maximum height

B = point of half maximum height

V_{A} = V_{0} \cos α

h_{A} = \frac{V_{0}^{2} \sin^{2} α}{2 g}

V_{A} = \sqrt{\frac{2}{5}} V_{B} \Rightarrow V_{B} = \frac{5}{2} \sqrt{\frac{2}{5}} V_{A} = \frac{5}{2} \sqrt{\frac{2}{5}} V_{0} \cos α

h_{B} = \frac{1}{2} h_{A} = \frac{V_{0}^{2} \sin^{2} α}{4 g}

{KE}_{A} + {PE}_{A} = {KE}_{B} + {PE}_{B}

\frac{1}{2} m . V_{A}^{2} + m . g . h_{A} = \frac{1}{2} m . V_{B}^{2} + m . g . h_{B}

\frac{1}{2} m {(V_{0} \cos α)}^{2} + m . g . \frac{V_{0}^{2} \sin^{2} α}{2 g} = \frac{1}{2} m {(\frac{5}{2} \sqrt{\frac{2}{5}} V_{0} \cos α)}^{2} + m . g . \frac{V_{0}^{2} \sin^{2} α}{4 g}

\frac{1}{2} \cos^{2} α + \frac{1}{2} \sin^{2} α = \frac{5}{4} \cos^{2} α + \frac{1}{4} \sin^{2} α

\frac{1}{4} \sin^{2} α = \frac{3}{4} \cos^{2} α

\tan^{2} α = 3 \Rightarrow \tan α = \sqrt{3} \Rightarrow α = 60 °

Commented by Tinkutara last updated on 23/Jun/17

But answer is 60 ° .

Commented by sandy_suhendra last updated on 23/Jun/17

sorry, I^{'} ve a mistake at the second line from the bottom

but I^{'} ve corrected

Commented by Tinkutara last updated on 24/Jun/17

Thanks Sir!

Answered by ajfour last updated on 23/Jun/17

A t h a l f t h e m a x . h e i g h t :

\frac{1}{2} {m v}^{2} = \frac{1}{2} {m u}^{2} - \frac{m g H}{2}

v^{2} = u^{2} - \frac{u^{2} \sin^{2} θ}{2} \dots (i)

A l s o a s g i v e n u \cos θ = \sqrt{\frac{2}{5}} v

o r v^{2} = \frac{5}{2} u^{2} \cos^{2} θ (i i)

e q u a t i n g (i) w i t h (i i)

u^{2} - \frac{u^{2} \sin^{2} θ}{2} = \frac{5}{2} u^{2} \cos^{2} θ

(1 + \tan^{2} θ) - \frac{1}{2} \tan^{2} θ = \frac{5}{2}

\frac{1}{2} \tan^{2} θ = \frac{3}{2} \Rightarrow \tan θ = \pm \sqrt{3}

θ = 60 °, 120 ° .

Commented by Tinkutara last updated on 24/Jun/17

Thanks Sir!