Question

A particle is projected with speed u at angle theta with horizontal from ground if it is it same height from ground at time t1 and t2 then its average velocity and time interval t1 to t2 is

Answer 1

when particle is projected with some speed u and at an angle with horizontal

then its component of velocity will be

$u_y = usin\theta$

$u_x = ucos\theta$

now we have to find the average velocity when it comes back to same height at two time instants i.e. t1 and t2

so while it is at same time instant the displacement in vertical direction is ZERO while displacement in horizontal direction will not be zero

so its average velocity will be the average horizontal velocity

Now since we know that in horizontal direction the velocity will remain the same and it will not change so average velocity will same as initial horizontal velocity

$v_{avg} = ucos\theta$

now in order to find the time interval t1 and t2 we know that its height is H

$H = usin\theta * t - \frac{1}{2}gt^2$

$\frac{1}{2}gt^2 - usin\theta*t + H = 0$

now since its a quadratic equation

$t_1 + t_2 = \frac{2usin\theta}{g}$

$t_1t_2 = \frac{2H}{g}$

so here we have

$t_1 - t_2 = \sqrt{(t_1+t_2)^2 - 4t_1t_2}$

$t_1 - t_2 = \sqrt{(\frac{2usin\theta}{g})^2 - \frac{8H}{g}}$

so above is the time interval of the given instants