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

The baseball reaches a maximum height of 27 ft. The equation to represent the position of the baseball above the ground at a given time is p(t) = 1/2gt^2 + v0t + p0, where g is the acceleration due to gravity, v0 is the initial velocity, and p0 is the initial position above the ground. By finding the vertex of the parabolic function, we can determine the maximum height reached by the ball.

Step-by-step explanation:

The equation to represent the position of the baseball above the ground at a given time is p(t) = 1/2gt^2 + v0t + p0.

In this equation, g is the acceleration due to gravity which is -32 ft/sec^2 and t is the time in seconds.

The initial velocity, v0, is given as 24 ft/sec and the initial position, p0, is given as 5 ft above the ground.

To find the maximum height reached by the ball, we need to determine the vertex of the parabola represented by the quadratic function.

The vertex of a parabola can be found using the formula t = -v0/(2g), where t is the time at which the maximum height is reached.

Plugging in the given values, we get t = -24/(2*(-32)) = 3/4 sec.

To find the maximum height, substitute the value of t into the quadratic function.

p(3/4) = 1/2*(-32)*(3/4)^2 + 24*(3/4) + 5

= 27 ft.

Therefore, the baseball reaches a maximum height of 27 ft.