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A ball is thrown into the air where its height in feet (h) and time in seconds (t) can be found by the following equation, h = -16t^2 + 48t + 288. Find the maximum height of the ball and how long it takes for the ball to hit the ground.

a. Maximum height is 336 feet, and it takes 3 seconds to hit the ground.
b. Maximum height is 288 feet, and it takes 2 seconds to hit the ground.
c. Maximum height is 288 feet, and it takes 3 seconds to hit the ground.
d. Maximum height is 336 feet, and it takes 2 seconds to hit the ground.

User Kaulex
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1 Answer

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Final answer:

The maximum height of the ball is 336 feet, reached at 1.5 seconds. The ball takes 6 seconds to hit the ground, not 3 seconds as might be initially assumed if the intercepts with the height axis are not considered.

Step-by-step explanation:

To determine the maximum height of the ball and the time it takes for the ball to hit the ground when it is thrown into the air according to the equation h = -16t^2 + 48t + 288, we must first understand that this is a quadratic equation in the form at^2 + bt + c, where a, b, and c are constants. To find the time at which the height is maximum, we use the formula for the vertex of a parabola, which is -b/(2a). Plugging in our values gives us the time t at maximum height, t = -48/(2*(-16)) = 1.5 seconds. Plugging this back into the height equation gives us the maximum height, h = -16*(1.5)^2 + 48*(1.5) + 288 = 336 feet.

To find the time it takes for the ball to hit the ground, we set the height equation equal to zero, 0 = -16t^2 + 48t + 288, and solve for t. Factoring or using the quadratic formula yields t = 3 seconds and t = 6 seconds. Since the question assumes the ball is thrown from the ground (height = 0), we take the larger value which is when the ball hits the ground on its way down, hence t = 6 seconds.

User Karimah
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