During a baseball game, a player hits a a ball with a speed of 43m/s at an angle of 25∘ above the horizontal. When the player hit the ball, it was 1m above the ground, and after the hit, the ball flies straight toward the center field fence.

How high above the ground is the ball when it reaches the center field fence, which is a distance of 400ft (122m) away?

s_y = 9.82 m

Explanation:

Given:

- Initial velocity v_i = 43 m/s

- Angle with the horizontal Q = 25 degree

- Initial distance s_o = 1 m

- The distance of the center field fence x_f = 122 m

Find:

- How high above the ground is the ball when it reaches the center field fence

Solution:

- The time taken for the ball to reach the fence t_f:

s_x = S (0) + v_x, o*t

122 = 0 + (43*cos (25)) * t

t = 122 / (43*cos (25)) = 3.1305 s

- Compute the height of the ball when it reaches the fence:

s_y = S (0) + v_y, o*t + 0.5*g*t^2

s_y = 1 + 43*sin (25) * 3.1305 - 0.5 * (9.81) * (3.1305) ^2

s_y = 9.82 m