Physics - Kinematics Question with Solution | TestHub

The height $h$ of the ball at time $t$ is given by $h=vt-\frac{1}{2}g{t}^2$.

Given $v=80ft/sec$ and $h=96\mathrm{ft}$. Using $g=32ft/se{\mathrm{c}}^2$:

$96=80t-\frac{1}{2}(32)t^2$

$96=80t-16t^2$

$16t^2-80t+96=0$

Divide by 16:

$t^2-5t+6=0$

Factoring the quadratic equation:

$(t-2)(t-3)=0$

Thus, $t=2\sec$ or $t=3\sec$.