Question 890666
When a basketball player shoots a ball from his hand at an initial of 2 m with an initial upward velocity of 10 meters per second, the height of the ball can be modeled by the quadratic expression -4.9t^2 + 10t + 2 after t seconds.
a. What will be the height of the ball after 2 seconds?
h(t) = -4.9t^2 + 10t + 2
Sub 2 for t
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b. How long will it take the ball to reach the height of 4.5 m?
h(t) = -4.9t^2 + 10t + 2 = 4.5
Solve for t
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  How long will it take to touch the ground?
h(t) = -4.9t^2 + 10t + 2 = 0
Solve for t
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c. Do you think the ball will reach the height of 12 m?
The max height is the vertex of the parabola.
The vertex is on the axis of symmetry, t = -b/2a = -10/-9.8
Max ht is at 50/49 seconds
h(t) = -4.9t^2 + 10t + 2
Sub (50/49) for t for max ht.
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 Why?
d. Will the ball hit the ring if the ring is 3 m high?
Depends on the max ht above.
h(t) give the height of the center of the ball, take that into account.
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e. Write a similar situation with varied initial height when the ball is thrown with an initial upward velocity. Then model the path of the ball by quadratic expression.
h(t) = -4.9t^2 + 10t + ho (ho = initial height)
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f. Using the equation and the quadratic expression you have written in item e, formulate and solve problems involving the height of the ball when it is thrown after a given time.
Same as above