Question

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

Answer 1

Topic :- Motion in plane

$\maltese \: \underline{\textbf{\textsf{AnsWer}}}\:\maltese$

• Time of flight (T) = 3 sec
• Horizontal Range (R) = 150 m
• Velocity (u) = ?

It's given that, the time of flight for that range is 3

sec :

$\longrightarrow\:\:\sf Time \: of \: flight = 3 \: sec$

$\qquad\footnotesize\bullet\:\sf Time \: of \: flight = \dfrac{2u_{y}}{g} = \dfrac{2u \sin(\theta)}{g}$

$\longrightarrow\:\:\sf \dfrac{2u \sin( \theta) }{g} = 3 \: sec$

Taking Acceleration due gravity (g) as 10 m/s² we have :

$\longrightarrow\:\:\sf \dfrac{2u \sin( \theta) }{10} = 3 \: sec$

$\longrightarrow\:\:\sf \dfrac{u \sin( \theta) }{5} = 3$

$\longrightarrow\:\:\sf u \sin( \theta) = 3 \times 5$

$\longrightarrow\:\:\sf u \sin( \theta) = 15 \qquad \:....(e {q}^{n} \: i)$

Now, we're also given that, horizontal range is 150m :

$\dashrightarrow\:\:\sf Horizontal \: Range = 150 \: m$

$\qquad\footnotesize\bullet\:\sf Horizontal \: Range = \dfrac{u^2 \sin2(\theta)}{g}$

$\dashrightarrow\:\:\sf \dfrac{u^2 \sin2(\theta)}{g} = 150$

$\qquad\footnotesize\bullet\:\sf \sin2(\theta) =2 \sin( \theta) . \cos( \theta)$

$\dashrightarrow\:\:\sf\dfrac{u^22 \sin( \theta).\cos( \theta)}{g} = 150$

$\qquad\footnotesize\bullet\:\sf u^22 \sin( \theta).\cos( \theta) =2u \sin( \theta).u \cos( \theta)$

$\dashrightarrow\:\:\sf\dfrac{2u \sin( \theta). u\cos( \theta)}{g} = 150$

$\dashrightarrow\:\:\sf\dfrac{2u \sin( \theta).u \cos( \theta)}{10} = 150$

$\dashrightarrow\:\:\sf\dfrac{u \sin( \theta).u \cos( \theta)}{5} = 150$

$\dashrightarrow\:\:\sf u \sin( \theta). u\cos( \theta)= 150 \times 5$

$\dashrightarrow\:\:\sf u\sin( \theta). u\cos( \theta)= 750$

$\qquad\footnotesize\bullet\:\sf Replacing \: u \sin( \theta) = 15 \: from \: equation \: (i) \: we \: have:$

$\dashrightarrow\:\:\sf 15. u\cos( \theta)= 750$

$\dashrightarrow\:\:\sf u\cos( \theta)= \dfrac{750}{15}$

$\dashrightarrow\:\:\sf u\cos( \theta)= 50$

Now, let's find the angle of projection first :

$:\implies \sf \dfrac{u\sin(\theta)}{u \cos (\theta)} = \dfrac{15}{50}$

$:\implies \sf \dfrac{\sin(\theta)}{ \cos (\theta)} = \dfrac{15}{50}$

$:\implies \sf \tan \theta = \dfrac{15}{50}$

$:\implies \sf \theta = \tan^{ - 1} \bigg(\dfrac{15}{50} \bigg)$

$:\implies \sf \theta = \tan^{ - 1} \bigg(0.3\bigg)$

$:\implies \sf \theta = 16.6 {9}^{ \circ}$

Now, put the value of θ = 16.69° in the eqⁿ (i) :-

$\longrightarrow\:\:\sf u \sin( \theta) = 15$

$\longrightarrow\:\:\sf u \sin( 16.6 {9}^{ \circ} ) = 15$

$\qquad\footnotesize\bullet\:\sf \sin(16.69^{ \circ}) \approx 0.30$

$\longrightarrow\:\:\sf u \times 0.30 = 15$

$\longrightarrow\:\:\sf u=\dfrac{15}{0.30}$

$\longrightarrow\:\:\sf u=\dfrac{15 \times 10}{0.30 \times 10}$

$\longrightarrow\:\:\sf u=\dfrac{150}{3}$

$\longrightarrow\:\:\sf u=\dfrac{150}{3}$

$\longrightarrow\:\: \underline{ \boxed{\sf u=50 {ms}^{ - 1} }}$

Answer 2

Answer:

answer is 52.20153254

Explanation:

the sine or cosine of the found angle you divide the number it is equated to it gives that answer from the two earlier equations we had