P01S2016 – Throwing a Rock from a Truck

Throwing a Rock from a Truck

Louise is standing on the back of a pickup truck driving down the road with speed $v_T=5\ m/s$, and she throws a rock up. Her friend Thelma is standing on the side of the road (see figure below), and she sees this rock go straight up to a height $h=9.8/8\ m$, and then fall straight down again.

1. What is the velocity ${\overrightarrow{v}}_0$ of the rock when it leaves Louise’s hand that Louise sees? (Once you specify your reference frame, you can give me the two components of the velocity or its magnitude and direction). You don’t have to take into account Louise’s height or the height of the truck.

For Thelma to see the rock thrown straight up into the air, Louise has to throw it at an angle such that its horizontal component $v_{0x}$ cancels with the truck’s speed $v_T$.

Thus, we conclude that the $x$ component $v_{0x}$ of the launch velocity must equal $v_{0x}=-5\ m/s$ in the frame of reference of Louise. In addition, its $y$ component $v_{0y}$ must be such that the rock rises to a height $h$ and it must therefore satisfy the following equation

\begin{aligned}
v^2_{fy}=v^2_{0y}+2a_y\mathrm{\Delta }y\ \ \ \ &\Rightarrow \ \ \ \ \ 0=v^2_{0y}-2gh \\
\\
&\Rightarrow \ \ \ \ \ v_{0y}=\sqrt{2gh}=\sqrt{2\cdot 9.8\cdot \frac{9.8}{8}}=4.9\ m/s
\end{aligned}

The rock must therefore have been thrown with a velocity ${\overrightarrow{v}}_0$ measured in Louise’s reference frame that is equal to

\boxed{{\overrightarrow{v}}_0=-v_T\ \hat{x}+\sqrt{2gh}\ \hat{y}=-5\ \hat{x}+4.9\ \hat{y}\ \ \ \ (m/s)}