# After 2 seconds of flight,the velocity of a projectile is (12i-4j)m/s,Find its speed and elevation at moment of projection?

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Feb 9, 2018

v_0=sqrt(v_x^2+(v_y+g t)^2
$\Delta y = {v}_{y} t + \frac{1}{2} g {t}^{2}$
where  (v_x,v_y)=(12,-4)m/s;, t=2s

#### Explanation:

Let's set t=0 at the moment of projection, and denote
${v}_{0} , {v}_{0 x} , {v}_{0 y}$ as initial velocity and initial velocities in x- and y- direction.

$\left({v}_{x} , {v}_{y}\right) = \left(12 , 4\right) \frac{m}{s}$

$\because {v}_{y} = {v}_{0 y} - g t$
$\Rightarrow {v}_{0 y} = {v}_{y} + g t$

Substitute the above into the y(t) equations

y=y_0+v_(0y)t-½g t^2
$\Delta y = \left({v}_{y} + g t\right) t - \frac{1}{2} g {t}^{2} = {v}_{y} t + \frac{1}{2} g {t}^{2}$

Because,
${v}_{0 x} = {v}_{0} \cos \theta = {v}_{x}$
${v}_{0 y} = {v}_{0} \sin \theta = {v}_{y} + g t$

rArr v_0= sqrt(v_(0x)^2+v_(0y)^2)=sqrt(v_x^2+(v_y+g t)^2

Substitute ${v}_{x} = 12 \frac{m}{s} , {v}_{y} = - 4 \frac{m}{s} , \mathmr{and} t = 2 s$, you should be able to calculate the elevation, and the initial velocity.

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Jane Share
Feb 9, 2018

When a projectile is projected with an initial speed of $u$ at an angle of $\theta$ w.r.t to horizontal,its horizontal component of velocity i.e $u \cos \theta$ remains constant through out.

But, the vertical component of velocity i.e $u \sin \theta$ decreases to zero,while going up,as gravitational force opposes that motion.

So,suppose,after $2 s$ of its journey, it had a vertical velocity of $v$ so we can write, $v = u \sin \theta - g \cdot 2$ (using, $v = u - g t$,here, initial velocity upwards is $u \sin \theta$)

So,in vector form we can write,velocity after $2 s$ as,

$\left(u \cos \theta\right) i + \left(u \sin \theta - 2 g\right) j$

So,comparing with the given equation,we can say,

$u \cos \theta = 12$ and $u \sin \theta - 2 g = - 4$

Solving the above two equations we get, $u = 20 \frac{m}{s}$ and $\theta = {\cos}^{-} 1 \left(\frac{3}{5}\right) = 53$ degrees

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