Question

Baseball player a bunts the ball by hitting it in such a way that it acquires an initial velocity of 2.4 m/s parallel to the ground. upon contact with the bat the ball is 1.2 m above the ground. player b wishes to duplicate this bunt, in so far as he also wants to give the ball a velocity parallel to the ground and have his ball travel the same horizontal distance as player a's ball does. however, player b hits the ball when it is 1.6 m above the ground. what is the magnitude of the initial velocity that player b's ball must be given

Answer 1

Let $\mathbf r_A$ denote the position vector of the ball hit by player A. Then this vector has components

$\begin{cases}r_{Ax}=\left(2.4\,\frac{\mathrm m}{\mathrm s}\right)t\\r_{Ay}=1.2\,\mathrm m-\frac12gt^2\end{cases}$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the magnitude of the acceleration due to gravity. Use the vertical component $r_{Ay}$ to find the time at which ball A reaches the ground:

$1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s$

The horizontal position of the ball after 0.49 seconds is

$\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m$

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector $\mathbf r_B$ of the ball hit by player B has

$\begin{cases}r_{Bx}=v_0t\\r_{By}=1.6\,\mathrm m-\frac12gt^2\end{cases}$

Again, we solve for the time it takes the ball to reach the ground:

$1.6\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.57\,\mathrm s$

After this time, we expect a horizontal displacement of 12 meters, so that $v_0$ satisfies

$v_0(0.57\,\mathrm s)=12\,\mathrm m$

$\implies v_0=21\,\dfrac{\mathrm m}{\mathrm s}$