Question

can I know the answer for this question?​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore Initial\:velocity=50\:m/s}}$

${\bold{\therefore Angle\:of\:projection=sin^{-1}(\frac{3}{10})}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a missile whose horizontal range and time of flight is given.

• We have to find the initial velocity for this projection and angle of projection.

$\underline \bold{Given : } \\ \implies Range = 150 \: m \\ \\ \implies Time \: of \: flight = 3 \: sec \\ \\ \implies Acceleration(a) = 10 \: m/ {s}^{2} \\ \\ \underline \bold{To \: find: } \\ \implies Initial \: velocity(u) = ? \\ \\ \implies Angle \: of \: projection = ?$

$\bold{Using \: formula \: of \: range : } \\ \implies R= \frac{ {u}^{2} \: sin \: 2\theta}{g} \\ \\ \implies 150 = \frac{ {u}^{2} \: sin \: 2 \theta }{10} \\ \\ \implies {u}^{2} \: sin \: 2 \theta = 1500 \\ \\ \implies sin \: 2 \theta = \frac{1500}{ {u}^{2} } \\ \\ \implies sin \theta = \frac{ \cancel{1500}}{ {u}^{2} \: \cancel2 } \\ \\ \implies sin \theta = \frac{750}{ {u}^{2} } - - - - - (1) \\ \\ \bold{Using \: formula \: of \: tim \: of \: flight : } \\ \implies T = \frac{2u \: sin \theta}{g} \\ \\ \implies 3 = \frac{2u \: sin \theta}{10} \\ \\ \bold{Putting \: value \: of \: sin \theta : } \\ \implies 3 = \frac{2 \times u \times 750}{ {u}^{\cancel2} \times \cancel10 } \\ \\ \implies 3u = 150 \\ \\ \implies u = \frac{\cancel{150}}{\cancel3} \\ \\ \bold{\implies u = 50 \:m/s} \\ \\ \bold{Putting \: value \: of \: u \: in \: (1)} \\ \implies sin \theta = \frac{750}{ {50}^{2} } \\ \\ \implies sin \theta = \frac{3}{10} \\ \\ \bold{\implies \theta = {sin }^{ - 1} (\frac{3}{10} })$

Answer 2

$\huge{\textbf{\underline{\green{Answer:-}}}}$

$\boxed{\bold{u = 50}}$

$\boxed{\bold{ \theta = sin^{-1}\frac{3}{10}}}$

$\huge{\textbf{\underline{\green{Step-by-Step Explanation:-}}}}$

Question:- find the velocity of projection and angle of projection of a missile which has a horizontal range of 150m, if the time of flight for that range in 3 second (g = 10 m/s²).

Solution:-

Given,

Range of the missile = 150m

Time of flight = 3s

Acceleration of flight = 10 m/s²

Here, we have to find out initial velocity of projection and angle of projection of a missile.

By using the formula of Range,

$\boxed{R = \frac{u^2 sin 2\theta}{g}}$

$\implies{150 = \frac{u^2 sin 2\theta}{10}}$

$\implies{u^2 sin 2\theta = 1500}$

$\implies{ sin 2\theta = \frac{1500}{u^2}}$

$\implies{ sin \theta = \frac{1500}{u^2 × 2}}$

$\implies{ sin \theta = \frac{750}{u^2} →(Equation 1)}$

By using the formula of time of flight,

$\boxed{T = \frac{2u sin \theta}{g}}$

$\implies{3 = \frac{2u sin \theta}{10}}$

Putting the given values,

$\implies{3 = \frac{2 × u × 750}{u^2× 10}}$

$\implies{3u = 150}$

$\boxed{\bold{.°. u = 50}}$

Now,

Putting the given values in equation (1),

$\implies{ sin \theta = \frac{750}{u^2}}$

$\implies{ sin \theta = \frac{750}{50^2}}$

$\implies{ sin \theta = \frac{750}{2500}}$

$\implies{ sin \theta = \frac{3}{10}}$

$\boxed{\bold{ .°.\theta = sin^{-1}\frac{3}{10}}}$