Question

A baseball bat makes contact with a ball 0.65 meters above the ground and the ball moves away at a 43 degree angle from the horizontal. it's a homer!! The ball lands in the bleachers (the seats) 128.5 meters away from home plate, 4.3 meters above the ground. How long was the ball in the air? How fast was the ball going when it left the bat? The ball was in the air for The ball was moving at a speed of m/s when it left the bat

Answer 1

The ball was in the air for approximately 3.21 seconds and it was moving at a speed of approximately 40.06 m/s when it left the bat.

Step-by-step explanation:

To calculate the time of flight, we can use the vertical motion equation: d = v*t + 0.5*a*t^2. Since the ball lands at the same height it was hit, the vertical displacement is zero. We can rearrange the equation to solve for time: t = -v_y/a. Substituting the values, we have t = -43*sin(43)/(-9.8) ≈ 3.21 seconds.

To calculate the initial velocity, we can use the horizontal motion equation: d = v*t. Rearranging the equation to solve for velocity, we have v = d/t. Substituting the values, we have v = 128.5/3.21 ≈ 40.06 m/s.