Question

A baseball player hits a high pop-up with an initial velocity of 32 meters per second from a height of 1.3 meters above the ground. The height, h(t), of the ball in meters, t seconds after being hit is modeled by the equation: h(t) = -4.9t² + 32t + 1.3. What is the maximum height reached by the ball, and how long does it take for the ball to reach its maximum height?

Answer 1

The baseball reaches a maximum height of 52.84 meters after approximately 3.27 seconds.

Step-by-step explanation:

To find the maximum height reached by the ball and the time it takes to reach that height, we use the given quadratic equation for height, h(t) = -4.9t² + 32t + 1.3, which models the vertical motion of the baseball. The maximum height is achieved when the derivative of this equation, which represents the velocity, is equal to zero. This happens at the peak of the trajectory when the ball momentarily stops ascending and begins to descend.

The derivative of the height equation with respect to time t is -9.8t + 32. Setting this equal to zero to find when the velocity is zero (maximum height), we solve -9.8t + 32 = 0 to get t = 32 / 9.8, which is approximately 3.27 seconds.

Substitute t = 3.27 back into the original height equation to find the maximum height: h(3.27) = -4.9(3.27)² + 32(3.27) + 1.3. This simplifies to approximately 52.84 meters.

Thus, the maximum height reached by the ball is 52.84 meters, and it takes 3.27 seconds to reach this height.