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A ball is thrown directly upward from ground level with an initial speed of 80ft/s. How high will it go? When will it return to the ground? use the formula h = vt - 16t²

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Final answer:

The maximum height reached by the ball thrown upward at 80ft/s is 100 feet, achieved in 2.5 seconds. The ball will return to the ground after 5 seconds total flight time.

Step-by-step explanation:

The question asks how high a ball thrown directly upward with an initial speed of 80ft/s will go and when it will return to the ground using the equation h = vt - 16t². To find the maximum height, we need to find the time when the velocity is zero (the peak of the trajectory). The velocity at any time t is given by v = 80 - 32t (from the equation of motion v = u + at, where a is the acceleration due to gravity, here taken as -32 ft/s² as there are two 16s in 32). Setting this to zero gives a time of t = 2.5 s at the peak.

Plugging t = 2.5 s into the original equation yields the maximum height: h = 80(2.5) - 16(2.5)² = 200 - 16(6.25) = 200 - 100 = 100 ft.

For the time to return to the ground, the height h will be 0. Solving 0 = 80t - 16t² yields t = 0 or t = 5 s. Ignoring the initial time t = 0, the ball will take 5 seconds to hit the ground.

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