Question

A baseball is thrown in a parabolic arc. It's position above the ground at a given point in time can be represented by the quadratic function p ( t ) = 1 2 g t 2 + v 0 t + p 0 , where t ≥ 0 , g is -32 ft/sec/sec, v 0 is initial velocity, and p 0 is its initial position above the ground. If the ball was thrown straight up at 24 ft/sec when it was 5 ft above the ground, how high did it go?

Answer 1

To find the maximum height the baseball reached, we first calculated the time at the vertex of the parabola to be 0.75 seconds. Then we found the maximum height by substituting this time into the position equation, which gave us a result of 16.25 feet above the ground.

Step-by-step explanation:

To determine how high the baseball went after being thrown straight up with an initial velocity of 24 ft/sec from a height of 5 ft above the ground, we can use the quadratic function for its position over time, p(t) = (1/2)gt² + v0t + p0. Here, g is the acceleration due to gravity (-32 ft/sec²), v0 is the initial velocity, and p0 is the initial position.

To find the maximum height, we need to determine the time at which the vertical velocity is zero. This occurs at the vertex of the parabola represented by the quadratic equation.

The formula to find the time at the vertex is tv = -v0/g.

Plugging in our values, we get tv = -24 / (-32), which simplifies to tv = 0.75 sec.

We then use this time to find the maximum height by substituting tv back into the position function:

p(tv) = (1/2)(-32)(0.75)2 + 24(0.75) + 5.

Simplifying gives us p(0.75) = -12(0.5625) + 18 + 5, which gives us p(0.75) = -6.75 + 18 + 5, resulting in p(0.75) = 16.25 ft. Thus, the maximum height reached by the baseball is 16.25 feet.