a ball is thrown from the top of a 72 ft building with a upward velocity of 24ft/s

Question

A.when will it reach its maximum height

B.how far above the ground will it be

C.how long will it take for the ball to reach the ground

formula h=-16t^2+v t+h

Answer 1

Hi Jay, the equation to solve represents a parabola. The highest point with be the vertex, represented by:

(t, h)

Here t represents the time when the ball gets to its maximum height and h represents the maximum height.

The vertex of a parabola can be found by first finding t = -b / 2a.

Here b = 24 and a = -16.

t = -24 / (2 * -16), or -24 / -32 = 3 / 4 seconds

The height at time 3 / 4 seconds is the value of the function: h(3/4) = -16 * (3 / 4)² + 24 * 3 / 4 + 72

h(3 / 4) = 81 feet

The ball will reach its maximum height 3 / 4 seconds after being thrown.

The ball will reach a maximum height of 81 feet.

To find how long the ball is in the air, solve the equation: -16t² + 24t + 72 = 0

Start by dividing both sides by 8:

-2t² + 3t + 9 = 0

This can be factored several ways, here is one solution: (t - 3) * (-2t - 3)

Using the zero product rule:

t - 3 = 0

t = 3

And

-2t - 3 = 0

t = -3 / 2, this solution can be discarded because it is negative.

The ball will hit the ground 3 seconds after it was thrown.

Questions?