Answer :
Final answer:
The initial speed of the ball is approximately 43.6 ft/s because it was hit and caught at the same height, allowing us to use kinematic equations to calculate its velocity.
Therefore, the initial speed of the ball is approximatelyB) 43.6 ft/s.
Explanation:
To find the initial speed of the ball, we can use the kinematic equations of motion. Given that the ball was hit 2.5 feet above the ground and caught by an outfielder 2 feet above the ground, the change in vertical displacement is Δy = 2 ft - 2.5 ft = -0.5 ft. Using the equation[tex]v_iy * t + (1/2) * g * t^2,[/tex] where v_iy is the initial vertical velocity, g is the acceleration due to gravity [tex](-32.2 ft/s^2),[/tex], and t is the time of flight, we can solve for v_iy.
Since the ball was hit and caught at the same height, the time of flight can be found using the equation Δy = (1/2) * g * [tex]t^2.[/tex] Substituting the given values, we find that t = sqrt((2 * 0.5) / 32.2) = 0.252 seconds. Now, we can use the equation[tex]v_iy[/tex]= (Δy - (1/2) * g * [tex]t^2)[/tex] t to find the initial vertical velocity. Substituting the values, we get [tex]v_iy[/tex] = (-0.5 - (1/2) * (-32.2) * [tex](0.252)^2)[/tex] / 0.252 = -43.6 ft/s. Since velocity is a vector quantity, the magnitude of the initial velocity is the absolute value of viy, which is 43.6 ft/s.
Therefore, the initial speed of the ball is approximately 43.6 ft/s.
