Chapter 9 - Linear Momentum and Collisions - Conceptual Questions - Page 189: 5

$1.38m$

We know that $p=mv$ $\implies v=\frac{p}{m}$ $v=\frac{0.780}{0.150}=5.20\frac{m}{s}$ We also know that: $y_{\circ}=\frac{v^2}{2g}$ We plug in the known values to obtain: $y_{\circ}=\frac{(5.20)^2}{2(9.81)}$ $y_{\circ}=1.38m$