Question

A particle is projected with velocity u at angle theta from horizontal find the magnitude of average velocity of particle in the duration in which it reaches the highest point of trajectory from the point of projection

Answer 1

Answer:

(v))($\sqrt{\frac{(1+cos^{2}x )}{2} }$)

Explanation:

average velocity=net displacement per unit time.

time=vsinФ/g

in that time, x displacement is (vsinФ/g)*vcosФ

y displacement =0.5(g)(VsinФ/g)^2

vector addition of x and y displacements:(v^2/g)(sinФ)($\sqrt{\frac{(1+cos^{2}x )}{2} }$)

Now displacement/ time.

=(v))($\sqrt{\frac{(1+cos^{2}x )}{2} }$)