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Imagine an astronaut is exploring another planet. He throws an object up in the air, and its height (in feet) upward from the surface of the planet is given by the function s(t) = −8.05t2 + 35t, where t is the time in seconds after the object is thrown.

User Fake Jim
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a) What is the initial height of the object?

The initial height of the object is the height at t = 0. Substituting t = 0 in the equation for s(t), we get:

s(0) = -8.05(0)^2 + 35(0) = 0

Therefore, the initial height of the object is 0 feet, which means it is on the surface of the planet.

b) What is the maximum height reached by the object?

The maximum height reached by the object occurs at the vertex of the parabolic function s(t). The vertex of a parabola of the form ax^2 + bx + c is given by the formula:

t = -b/2a

In this case, a = -8.05 and b = 35. Substituting these values, we get:

t = -35/(2*(-8.05)) = 2.17 seconds (rounded to two decimal places)

To find the maximum height, we substitute t = 2.17 seconds in the equation for s(t):

s(2.17) = -8.05(2.17)^2 + 35(2.17) = 34.09 feet (rounded to two decimal places)

Therefore, the maximum height reached by the object is 34.09 feet.

c) When does the object hit the ground?

The object hits the ground when its height s(t) is equal to 0. We can solve the equation -8.05t^2 + 35t = 0 to find the times when s(t) = 0:

-8.05t^2 + 35t = 0

t(-8.05t + 35) = 0

t = 0 or t = 4.34 seconds (rounded to two decimal places)

The first solution, t = 0, corresponds to the initial position of the object, so we can discard it. Therefore, the object hits the ground after approximately 4.34 seconds.

User Unickq
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