# A particle is projected from ground with speed 80m/s at an angle 30° with horizontal from ground.What is the magnitude of average velocity of particle in time interval t=2s to t=6s?

##### 1 Answer
Mar 18, 2018

Let's see the time taken by the particle to reach maximum height,it is, $t = \frac{u \sin \theta}{g}$

Given,$u = 80 m {s}^{-} 1 , \theta = 30$

so,$t = 4.07 s$

That means at $6 s$ it already started moving down.

So,upward displacement in $2 s$ is, $s = \left(u \sin \theta\right) \cdot 2 - \frac{1}{2} g {\left(2\right)}^{2} = 60.4 m$

and displacement in $6 s$ is $s = \left(u \sin \theta\right) \cdot 6 - \frac{1}{2} g {\left(6\right)}^{2} = 63.6 m$

So,vertical dispacement in $\left(6 - 2\right) = 4 s$ is $\left(63.6 - 60.4\right) = 3.2 m$

And horizontal displacement in $\left(6 - 2\right) = 4 s$ is $\left(u \cos \theta \cdot 4\right) = 277.13 m$

So,net displacement is $4 s$ is $\sqrt{{3.2}^{2} + {277.13}^{2}} = 277.15 m$

So,average velcoity = total displacement /total time=$\frac{277.15}{4} = 69.29 m {s}^{-} 1$