Question

Show that there are two values of time for which a projectile is at the same height. Also show that the sum of these two heights is equal to the time of flight

Answer 1

We know that the height of the projectile is given by:
$y=usin\theta t-\frac{1}{2}g{t}^2$
This equation can be rewritten for a particular height $y_1$ as:
$g{t}^2-2usin\theta t+2y_1=0$
The solution for the same would be given by:
$t=\frac{usin\theta \pm \sqrt{u^{2}si{n}^2\theta -2g{y}_1}}{g}$
Thus, the sum of these two times gives the time of flight, i.e., $\frac{2usin\theta }{g}$