Final answer:
Using principles of projectile motion and trigonometry, we calculated that the softball was hit with an initial speed of approximately 11.8 m/s at an angle of roughly 39.2 degrees above the horizontal.
Step-by-step explanation:
To determine the initial speed v0 and angle θ at which the softball was hit, we can use the principles of projectile motion. Since the third baseman catches the ball 2 seconds later and at the same height, we can ignore vertical motion for calculating v0 because vertical displacement is zero, and thus the vertical component of initial velocity is not required for v0 calculation. However, we need the angle for a complete answer. Given that the horizontal displacement L is 18.0m, we can calculate v0 using the formula for horizontal motion:
Horizontal displacement (L) = horizontal velocity (v0 * cos(θ)) * time (t)
We know that the horizontal velocity must be the component of the initial velocity in the direction the third baseman is running, which we can determine as follows:
L = v0 * cos(θ) * t
18.0m = v0 * cos(θ) * 2.00s
So, v0 * cos(θ) = 9.00 m/s.
To find the angle, we use the fact that the third baseman ran at a constant velocity of 7.00 m/s and caught the ball 2.00s later. Therefore, he ran a total distance of:
Distance run by third baseman = V * t = 7.00 m/s * 2.00 s = 14.0 m
Knowing the position where the ball was caught relative to where it was hit, we can then calculate the angle using trigonometry:
cos(θ) = 14.0m / 18.0m = 0.7778
Finally, θ = cos-1(0.7778) ≈ 39.2°.
Returning to the formula for v0, we get:
v0 = 9.00 m/s / cos(39.2°)
v0 ≈ 11.8 m/s
Thus, the initial speed of the softball was approximately 11.8 m/s, and it was hit at an angle of roughly 39.2 degrees above the horizontal.