## Physics (10th Edition)

The speed with which the pebble will hit the ground is $28.35m/s$.
a) When the pebble is fired horizontally, initially it has only horizontal velocity, so $v_{x}=14m/s$ and $v_{y0}=0$. $v_x$ does not change after that. On the vertical, the pebble then will start falling, corresponding to an increase in KE and a decrease in PE, so $$\Delta KE = -\Delta PE$$ $$\frac{1}{2}m\Delta v^2=-mg\Delta h$$ $$\frac{1}{2}\Delta v^2=-g\Delta h$$ We have $g=9.8m/s^2$, $\Delta h=-31m$ and $\Delta v=v_{yf}^2-v_{y0}^2=v_{yf}^2$ $$\frac{1}{2}v_{yf}^2=303.8$$ $$v_{yf}=24.65m/s$$ The speed with which the pebble will hit the ground is $v=\sqrt{v_{x}^2+v_{yf}^2}=28.35m/s$ For b) and c), here we are only concerned with conservative forces, meaning $\Delta KE$ and $\Delta PE$ are constant no matter how the pebble reaches the ground. Therefore, the speed with which the pebble will hit the ground is always the same, $v=28.35m/s$