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The height (in feet) of a ball thrown straight up into the air is given by h(t) = -16t² + 48t + 5. After how many seconds will the height of the ball be 37 feet when the ball is travelling towards the ground?

A 1
B 2
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D 4

User Dagang Wei
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1 Answer

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Final answer:

The height of the ball will be 37 feet when the ball is travelling towards the ground after approximately 2 seconds.

Step-by-step explanation:

To find the time it will take for the ball to be at a height of 37 feet when it is travelling towards the ground, we need to set the height function equal to 37 and solve for t:

-16t² + 48t + 5 = 37

-16t² + 48t - 32 = 0

We can solve this quadratic equation using the quadratic formula: t = (-b ± √(b² - 4ac)) / (2a). In this case, a = -16, b = 48, and c = -32.

Plugging in these values, we get: t = (-48 ± √(48² - 4(-16)(-32))) / (2(-16)).

Simplifying further, we have: t = (-48 ± √(2304 - 2048)) / (-32).

This simplifies to: t = (-48 ± √256) / (-32).

Since we're interested in when the ball is travelling towards the ground, we take the positive value of t: t = (-48 + 16) / (-32). Solving, we find that t is approximately 2 seconds.

User Shay Altman
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