Answer
1. Minimum speed: $v_x=26.3\frac{m}{s}$
2. Time in air: $t=0.714 s$
3. Distance traveled: $\Delta x=18.8m$
Work Step by Step
1. $\Delta y=y_{0}+v_{0y}\times t+\frac{1}{2}at^2$
$0.9m=2.50m+0\times t-4.9\frac{m}{s^2}\times t^2$
$t=\sqrt{\frac{1.6m}{4.9\frac{m}{s^2}}}=0.571s$
Minimum speed is: $v_x=\frac{15m}{0.571s}=26.3\frac{m}{s}$
2. Calculate time in air using same equation
$0m=2.5m-0\frac{m}{s} \times t-4.9\frac{m}{s^2}\times t^2$
$t=0.714 s$
3. To solve for distance traveled: $v_{x}t=\Delta x$
$\Delta x=26.3\frac{m}{s}\times 0.714s=18.8m$
