Question

figure 1 of 1 (figure 1)a softball is hit over a third basemans head with speed v0 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 horizontal - motion of the softball is a uniform - motion. The horizontal velocity \(v_x\) is constant. The third baseman runs with a constant velocity \(V = 7.00\ m/s\) and catches the ball at time \(t = 2.00\ s\). The initial horizontal position of the ball is \(x_0 = 0\) (starting from the point where it leaves the bat). The horizontal displacement \(x\) is given by \(x=v_x t\). Since the third baseman runs a distance \(x\) in time \(t\) with speed \(V\), and assuming the ball and the third - baseman have the same horizontal displacement when the ball is caught, \(x = Vt\). Substituting \(V = 7.00\ m/s\) and \(t = 2.00\ s\), we get \(x=(7.00\ m/s)\times2.00\ s = 14.0\ m\).

Step2: Analyze vertical motion

The vertical - motion of the softball is a uniformly - accelerated motion with acceleration \(a=-g=- 9.8\ m/s^{2}\) (taking up as positive). The initial vertical position \(y_0 = 0\) (starting from the height of the bat). The vertical displacement \(y\) is given by the equation \(y = y_0+v_{0y}t-\frac{1}{2}gt^{2}\). We are not given the initial vertical velocity \(v_{0y}\), but we can also use the fact that we want to find the position at \(t = 0.100\ s\). Using \(y = v_{0y}t-\frac{1}{2}gt^{2}\), assuming the ball starts at \(y = 0\). First, we need to consider the general vertical - motion formula \(y = v_{0y}t-\frac{1}{2}gt^{2}\). Since we are not given \(v_{0y}\), we can also consider the fact that the ball is in free - fall. At \(t = 0.100\ s\), \(y=0\times t-\frac{1}{2}(9.8\ m/s^{2})(0.100\ s)^{2}=- 0.0490\ m\).

Answer:

\(14.0,-0.0490\)