Question

A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 12.0 m/s. how long does he have to get out of the way if the shot was released at a height of 2.10 m, and he is 1.90 m tall?

Answer 1

The shot has a constant downward acceleration of $a_y=-9.81\,\dfrac{\mathrm m}{\mathrm s^2}$. Since acceleration is constant, the velocity of the ball is given by

$a_y=\dfrac{v_y-v_{0y}}t$

where $v_y$ is the shot's vertical velocity at time $t$ and $v_{0y}$ is its initial vertical velocity. We're given that $v_{0y}=12.0\,\dfrac{\mathrm m}{\mathrm s}$. So

$-9.81\,\dfrac{\mathrm m}{\mathrm s^2}=\dfrac{v_y-12.0\,\frac{\mathrm m}{\mathrm s}}t\implies v_y=12.0\,\dfrac{\mathrm m}{\mathrm s}+\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)t$

gives the shot's velocity at time $t$.

Because acceleration is constant, we also know that

$\bar v_y=\dfrac{d_y-d_{0y}}t=\dfrac{v_y+v_{0y}}2$

where $\bar v_y$ is the average vertical velocity, $d_y$ is the vertical displacement at time $t$, and $d_{0y}$ is the initial displacement at $t=0$. We're given an initial displacement of $d_{0y}=2.10\,\mathrm m$, so the displacement of the shot is

$d_y=d_{0y}+\dfrac12(v_y+v_{0y})t=d_{0y}+v_{0y}t+\dfrac12a_yt^2$

$\implies d_y=2.10\,\mathrm m+\left(12.0\,\dfrac{\mathrm m}{\mathrm s}\right)t+\dfrac12\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2$

The shotputter is 1.90 m tall, a difference of 0.20 m from the initial height of the throw. So we want to find the time for the ball to reach a displacment of -0.20 m:

$-0.20\,\mathrm m=2.10\,\mathrm m+\left(12.0\,\dfrac{\mathrm m}{\mathrm s}\right)t+\dfrac12\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2$

$\implies t=2.63\,\mathrm s$