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A ball is thrown into the air with an upward velocity of 13 ft/s. Its height h in feet after t seconds is given by the function h = −2t^2+ 13t + 24. In how many seconds does the ball reach its maximum height?

Answer :

Answer:

3.25 seconds

Step-by-step explanation:

The maximum height will occur when the velocity of the ball reaches zero in the air.

Velocity is given as the derivative of height, dh/dt.

The equation of height given is:

[tex]h = -2t^2+ 13t + 24[/tex]

Differentiating height, h, with respect to time, t, we have:

[tex]v = dh/dt = -4t + 13[/tex]

When v = 0:

[tex]0 = -4t + 13\\\\=> 4t = 13\\\\t = 13 / 4 = 3.25 secs[/tex]

Therefore, it will take 3.25 seconds for the ball to reach maximum height.

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