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Crystal is throwing a ball. The height at which the ball travels is given by a(t) = -16t^2 + 10t + 6 ft

When does the ball reach its maximum height?
When does the ball hit the ground?
What speed does the ball hit the ground at?

User Ymin
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To find when the ball reaches its maximum height, we can determine the vertex of the parabolic function representing the height of the ball. The vertex of a parabola in the form a(t) = at^2 + bt + c can be found using the formula t = -b / (2a).

In this case, a = -16 and b = 10. Substituting these values into the formula, we have:
t = -10 / (2*(-16))
t = -10 / (-32)
t = 5/16

Therefore, the ball reaches its maximum height at t = 5/16 seconds.

To find when the ball hits the ground, we need to determine the time when the height of the ball is zero. We can set a(t) = 0 and solve for t.

-16t^2 + 10t + 6 = 0

To solve this quadratic equation, we can factor it or use the quadratic formula. In this case, factoring is not straightforward, so we'll use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values a = -16, b = 10, and c = 6, we have:

t = (-10 ± √(10^2 - 4*(-16)6)) / (2(-16))
t = (-10 ± √(100 + 384)) / (-32)
t = (-10 ± √484) / (-32)
t = (-10 ± 22) / (-32)

We get two solutions: t = (-10 + 22) / (-32) and t = (-10 - 22) / (-32).

t = 12/(-32) = -3/8
t = -32/(-32) = 1

Since time cannot be negative in this context, we discard the negative solution. Therefore, the ball hits the ground at t = 1 second.

To find the speed at which the ball hits the ground, we can calculate the velocity at t = 1 second. The velocity function is the derivative of the height function, so we differentiate a(t) = -16t^2 + 10t + 6 with respect to t.

a'(t) = -32t + 10

Evaluating a'(t) at t = 1 second:

a'(1) = -32(1) + 10
a'(1) = -32 + 10
a'(1) = -22

Therefore, the speed at which the ball hits the ground is 22 ft/s (assuming downward direction as negative).
User Galloper
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