figure 1 of 1 (figure 1)a softball is hit over a third basemans head with speed v0 and at an angle θ from…

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)

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)

Answer

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)