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The function h(t) = -16t(t - 2) + 24 models the height h, in feet, of a ball t seconds after it is thrown straight up into the air. What are the initial velocity and the initial height of the ball? O 16 ft/s: 32 ft 32 ft/s: 24 ft 24 ft/s: 32 ft 48 ft/s: 24 ft

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Answer

Option B is correct.

Initial velocity = 32 ft/s

Initial height = 24 ft.

Step-by-step explanation

The function gives us the function that models the height, h, in feet for the ball as a function of time, t, in seconds.

h(t) = -16t (t - 2) + 24

We are then asked to find the initial velocity and the initial height of the ball.

This means we find the velocity and the height of the ball at t = 0

Velocity is given as the first derivative of the height function

v = (dh/dt)

h(t) = -16t (t - 2) + 24

h(t) = -16t² + 32t + 24

v = (dh/dt) = -32t + 32

When t = 0

v = -32t + 32 = -32 (0) + 32 = 0 + 32 = 32 ft/s

Initial velocity = 32 ft/s

For the initial height, t = 0

h(t) = -16t (t - 2) + 24

h(t) = -16t² + 32t + 24

h(0) = -16(0²) + 32(0) + 24

h(0) = 0 + 0 + 24

h(0) = 24 ft

Initial height = 24 ft.

Hope this Helps!!!

User Edgar Hernandez
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