Piece of Cake

Classical Mechanics Level 4

A particle is projected from the ground with an initial velocity of v at an angle θ \theta with the horizontal.The magnitude average velocity of the particle between its point of projection and the highest point of the trajectory is

v cos ⁡ θ v \cos \theta v 2 1 + 3 cos ⁡ 2 θ \frac{v}{2} \sqrt{1+3 \cos^2 \theta} v 2 1 + 2 cos ⁡ 2 θ \frac{v}{2} \sqrt{1+2 \cos^2 \theta} v 2 1 + cos ⁡ 2 θ \frac{v}{2} \sqrt{1+ \cos^2 \theta}

Shanthanu Rai by Shanthanu Rai , 21, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Shanthanu Rai
Apr 2, 2016

Average Velocity= d i s p l a c e m e n t t i m e \frac{displacement}{time}
v = H 2 + R 2 4 T 2 v= \frac{ \sqrt{H^2+ \frac{R^2}{4}}}{ \frac{T}{2}} ...(1)
Where, H=maximum height= v 2 sin ⁡ 2 θ 2 g \frac{v^2 \sin^2 \theta}{2g}
R=Range= v 2 sin ⁡ 2 θ g \frac{v^2 \sin 2\theta}{g}
T=time of flight= 2 v sin ⁡ θ g \frac{2v \sin \theta}{g}
Substituting in (1),
v = v 2 1 + 3 cos ⁡ 2 θ \boxed{v= \frac{v}{2} \sqrt{1 + 3 \cos^2 \theta}}

5 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...