Question

figure 1 of 1 (figure 1)a softball is hit over a third - basemans head with speed v and at an angle θ from the horizontal. immediately after the ball is hit, the third baseman turns around and runs straight back at a constant velocity v = 7.00 m/s, for a time t = 2.00 s. he then catches the ball at the same height at which it left the bat. the third baseman was initially l = 18.0 m from the location where the ball was hit at home plate. part d find a vector expression for the position (vec{r}) of the softball 0.100 s before the ball is caught. use the notation x, y, an ordered pair of values separated by a comma, where x and y are expressed numerically in meters, as measured from the point where the softball initially left the bat. express your answer to three significant figures. view available hint(s)

Explanation:

Step1: Analyze horizontal motion

The third - baseman runs with a constant velocity $v = 7.00\ m/s$ for $t = 2.00\ s$. The horizontal distance $x$ from the point where the ball was hit to the third - baseman is $L=18.0\ m$. The horizontal displacement of the ball in $t = 2.00\ s$ is $x_1$. The horizontal displacement of the third - baseman in $t = 2.00\ s$ is $x_{b}=vt=7.00\times2.00 = 14.0\ m$.

Step2: Find horizontal position of ball at $t = 2.00\ s$

The horizontal position of the ball at $t = 2.00\ s$ relative to the starting point (where it left the bat) is $x = L - x_{b}=18.0 - 14.0=4.00\ m$.

Step3: Analyze vertical motion (assuming free - fall, $y - y_0=v_{0y}t-\frac{1}{2}gt^{2}$, assume $y_0 = 0$ and we don't have information about initial vertical velocity to calculate $y$ exactly. But if we assume the ball is at the same height when caught as when hit, and we want the position $0.100\ s$ before being caught. Let's assume the motion is symmetric in the vertical direction around the mid - point of the flight. Since we have no information about vertical velocity, we assume the ball moves with constant horizontal velocity in the absence of air - resistance. In the last $0.100\ s$, the horizontal displacement $\Delta x=v_x\times0.100$. Since the horizontal velocity of the ball is the same as the average horizontal velocity of the motion related to the third - baseman's run. The horizontal position of the ball $0.100\ s$ before being caught is $x'=4.00 - v_x\times0.100$. Since the horizontal velocity of the ball is related to the third - baseman's motion, and assuming no acceleration in the horizontal direction, $v_x$ (horizontal velocity of the ball) is such that in $2.00\ s$ it covers the relative distance to the third - baseman. The horizontal position $x$ of the ball $0.100\ s$ before being caught: $x = 4.00-7.00\times0.100=3.30\ m$.

For the vertical position, since we have no information about initial vertical velocity and the ball is caught at the same height it was hit, we assume the ball is at the same height as when it was hit (mid - height of the trajectory in a symmetric case if no air - resistance). Let's assume the height of the ball when hit and caught is $y = 0$. So the position vector $\vec{r}=(3.30,0)$

Answer:

$3.30,0$