A-mango-in-a-tree-is-located-30-40-from-the-point-of-projection-of-stone-Find-the-minimum-speed-and-the-angle-of-projevtion-of-the-stone-so-as-to-hit-the-mango-

Question Number 18077 by virus last updated on 15/Jul/17

A m a n g o i n a t r e e i s l o c a t e d (30, 40) f r o m t h e

p o i n t o f p r o j e c t i o n o f s t o n e . F i n d t h e

m i n i m u m s p e e d a n d t h e a n g l e o f p r o j e v t i o n

o f t h e s t o n e s o a s t o h i t t h e m a n g o

Answered by ajfour last updated on 15/Jul/17

Equation of trajectory is

y = xtan θ - \frac{{gx}^{2}}{{2 u}^{2}} (1 + \tan^{2} θ)

point M (30, 40) must lie on it; so

40 = 30 \tan θ - \frac{10 (900)}{{2 u}^{2}} (1 + \tan^{2} θ)

differentiating and equating to zero :

0 = 30 \sec^{2} θ - \frac{9000}{{2 u}^{2}} (2 \tan θ \sec^{2} θ)

\Rightarrow \tan θ = \frac{{30 u}^{2}}{9000} = \frac{u^{2}}{300}

\Rightarrow 40 = 30 (\frac{u^{2}}{300}) - \frac{9000}{{2 u}^{2}} (1 + \frac{u^{4}}{90000})

\Rightarrow 40 = \frac{u^{2}}{10} - \frac{9000}{{2 u}^{2}} - \frac{u^{2}}{20}

40 = \frac{u^{2}}{20} - \frac{9000}{{2 u}^{2}} (. . \times {20 u}^{2} gives)

{800 u}^{2} = u^{4} - 90000

{(u^{2} - 400)}^{2} = 90000 + 160000

u^{2} - 400 = \sqrt{25 \times 10000_{}}

u^{2} = 900 \Rightarrow u_{\min} = 30 m / s,

θ = \tan^{- 1} (\frac{u^{2}}{300}) = \tan^{- 1} 3 .

Commented by virus last updated on 15/Jul/17

t h a n k y o u s i r

Commented by ajfour last updated on 15/Jul/17

Is answer correct, virus ?

Commented by virus last updated on 15/Jul/17

y e s s i r

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