Question

a particle is projected with speed u at angle theta with horizontal from ground. if it is at same height from ground at time t1 and t2, then it's average velocity in time interval t1 and t2 is

Answer 1

Final Answer : $u \cos(\theta) \: i$

Steps:
1) Average Velocity in from $t_1 \: to\: t_2$ = ?
We know,
It is given both are at same height at two times.

2)
Horizontal Component of velocity = $u \cos(\theta)$
Total Displacement from time $t_1 \:to\:t_2 = u \cos(\theta) *(t_1-t_2) i$
Total Time taken = $(t_1-t_2)$

3)
Average Velocity = Total Displacemnt / Total Time
=> Average Velocity = $u \cos(\theta) \: i$

Answer 2

Answer: Average velocity is $u.cos\theta$

To know the average time traveled from being one point of height to other which is same from the ground determines the parabolic shapes of projectile. Hence given as speed u at angle \theta and time as $t_1$ and $t_2$. Horizontal component of the velocity be u cos \theta and the total displacement in travelling from $t_1$ to $t_2$ be,

                     $u cos \theta \times ( t_1 -t_2)$  

=> Total time taken be $(t_1 -t_2)$.

Therefore the average velocity be,

                  $V_a = \frac {Total displacement} {total time}$

=> $Va = u cos \theta \times ( t_1 - t_2) / (t_1 - t_2) = u cos \theta.$