Question

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

Answer 1

Answer:

Let's see the time taken by the particle to reach maximum height,it is,  

t = u  sin θ \g

Given,

u = 80 m s∧ − 1 ,  θ= 30

so, t= 4.07 s

That means at 6 s  it already started moving down.

So,upward displacement in  2 s  is,  s = ( u sin θ )  ⋅  2 − 1  \2g(2) ∧2 = 60.4 m

and displacement in  6 s  is  s = (u  sin θ ) . 6 − 1\ 2 g ( 6 ) ∧2= 63.6 m

So,vertical dispacement in  ( 6 − 2)  =  4 s  is (63.6-60.4) = 3.2 m

And horizontal displacement in  ( 6 − 2 ) = 4 s  is  ( u  cos θ  ⋅  4 )  = 277.13 m

So,net displacement is  4 s  is  √ 3.2 ∧2 + 277.13 ∧2 = 277.15 m

So,average velcoity = total displacement /total time= 277.15 \4=69.29 ms∧-1

Explanation: