A ball was kicked from the ground into the air with an initial upward velocity of 32 feet per second. Write an equation to model the height of the ball and find the maximum height that it reached.

Question

answer the whole question

Answer 1

Hello Henry,

Your equation will represent the position of the ball with respect to time. I don't know if you have taken physics, but there is a physics equation that models this type of movement. The equation for this is...

Δx = v_ot + 0.5at²

Δx = distance from the ground

v_o = initial velocity

a = gravitational acceleration

You are given the initial velocity, and the acceleration due to gravity is 32 ft/s², so when you plug all these values in, you get...

Δx = 32t - 0.5(32)t²

Δx = 32t - 16t²

If you want it in the form y = ax² + bx + c, it would be as follows...

y = -16x² + 32x

This equation represent a parabola, and the vertex of this parabola is the time at which the ball was at its maximum height. The x value of the vertex of a parabola can be found using the formula -b/2a. In this case, a = -16 and b = 32, so...

-32/-(2)16 = 1

But 1 is not your final answer since this only gives you the time at which the ball got to its max height, but you still need to find the actual height, so just plug in 2 for 1 and solve for y.

y = -16(1)² + 32(1)

y = -16 + 32 = 16ft