find the velocity of projection of missile which has a horizontal rang

1.	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» 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) :- $\longrightarrow\:\:\sf u \sin( \theta) = 15 \\$ $\longrightarrow\:\:\sf u \sin( 16.6 {9}^{ \circ} ) = 15 \\$ $\qquad\footnotesize\bullet\:\sf \sin(16.69^{ \circ}) \approx 0.30\\$ $\longrightarrow\:\:\sf u \times 0.30 = 15 \\$ $\longrightarrow\:\:\sf u=\dfrac{15}{0.30} \\$ $\longrightarrow\:\:\sf u=\dfrac{15 \times 10}{0.30 \times 10} \\$ $\longrightarrow\:\:\sf u=\dfrac{150}{3} \\$ $\longrightarrow\:\:\sf u=\dfrac{150}{3} \\$ $\longrightarrow\:\: \underline{ \boxed{\sf u=50 {ms}^{ - 1} }} \\$

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»

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) :-

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

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

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

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

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

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

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

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