Question

find the velocity of projection of find the velocity of projection of missile which has horizontal range of 150m. if the time is a flight for that range is 3 seconds. ​

Answer 1

$\underline \textbf{ \large \: Given :- }$

$\star$ horizontal range of 150m.

$\star$ time is a flight for that range is 3 seconds.

$\underline \textbf{ \large \: Solution :- }$

We know that ,

time is a flight for that range is 3 seconds,

$\\ \: \: \: \: \: \: \: \: \: \: \: \sf \: T \: = \frac{2uy}{g} = \frac{2 \: u \: sin \theta}{g}$

$\\ \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \: \frac{2 \: u \: sin \theta \: }{g} = 3$

$\\ \: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: : \implies \: \frac{ \cancel2 \: ^{1} \: \: }{{ \cancel{10}} \: \: ^{5} } \: u \: sin \theta= 3$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: : \implies \boxed{\red{ \textbf{u \: sinθ \: = 15}}}.............................(1)$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \rm \overline{\underline{ horizontal\:range \: = \rm\frac{ u^{2} \: sin ^{2} \theta}{2g}} }$

$\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \rm : \implies \: \: \: \rm\frac{ u^{2} \: 2sin \theta \times cos \theta}{20} = \textbf{150}$

From eqⁿ 1 ,$\sf \: u \: = \frac{15}{sin \theta}$

225 × cos θ = 1500

cos θ = 1500 / 225 = 20/3

therefore , sin θ = 3 / √409

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