I have, what I think, is a relatively simple projectile motion problem. Yet I'm struggling to come up with a solution.
A particle is projected with a speed $u$ at and angle $\alpha$ to the horizontal. The particle takes 3 seconds to travel between 2 points $p$ and $q$ which are on the same horizontal line. Show the greatest height the particle reaches above this line is 11.025m
My attempt. Let's say the height above the ground of $p$ and $q$ is $h$. Let's also say the $t_1$ be the time to reach $p$ and $t_2$ be the time to reach $q$. Therefore, $t_2 - t_1 = 3$.
Using $$ s = ut + \frac{1}{2}at^2$$
I get $$ h = t_1u\sin\alpha - \frac{1}{2}gt_1^2$$ $$ h = t_2u\sin\alpha - \frac{1}{2}gt_2^2$$
where $u\sin\alpha = u_y$
I subtracted the second equation from the first, then simplified to get the result
$$2u\sin\alpha = g(t_1 + t_2) $$
I'm getting stuck since I have $h$, $u$ and $\sin\alpha$ as unknowns as well at $t_1$ and $t_2$.
Any help appreciated.
Think about this picture: The time the particle takes to go up above the pq horizontal line and reach the highest point is the same as the time it takes to come down back to the pq horizontal line, which is $\frac{3}{2}=1.5$ sec.
At the highest point, the vertical speed is $0$. Imagine you release it from that height, its initial vertical speed is 0, how far will it drop within 1.5 sec?
h=$\frac{1}{2}gt^2=\frac{1}{2}9.8(1.5)^2=11.025$.