Answer
(a) $2.10m/s$
(b) $0.07s$
Work Step by Step
(a) We can find the required launch speed as follows:
$v_y^2=v_{\circ y}^2-2a\Delta y$
At the maximum height $v_y=0$
$\implies 0=v_{\circ y}^2-2a\Delta y$
This simplifies to:
$v_{\circ y}=\sqrt{2g\Delta y}$
We plug in the known values to obtain:
$v_{\circ y}=\sqrt{2(9.8m/s^2)(0.22m)}$
$v_{\circ y}=2.1m/s$
(b) We can find the required time interval as follows:
$\Delta y=v_{\circ y}t+\frac{1}{2}gt^2$
As the initial velocity is zero $v_{\circ y}=0$
$\implies \Delta y=(0)t+\frac{1}{2}gt^2$
This simplifies to:
$t=\sqrt{\frac{2\Delta y}{g}}$
We plug in the known values to obtain:
$t=\sqrt{\frac{2(0.10m)}{9.8m/s^2}}$
$t=0.14s$
Thus, the interval between the successive stroboscopic exposures is half $0.14s$, that is $0.07s$.
