A basketball player throws a ball vertically with an initial velocity of 100ft/sec.The distance of the ball from the ground after t seconds is given by the expression 100t-16t^2.
1.what is the distance of the ball from the ground after 6 seconds?
2.after how many seconds does the ball reach a distance of 50 ft.from the ground?
3.how many seconds will it take for the ball to fall to the ground?
4.do you think the ball can reach the height of 160ft.?why?why not?
Answers
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Step-by-step explanation:
We calculate the maximum height of the ball, we find the vertex of the equation: ((-B / 2A),f(-B / 2A)) --> x = (-100/2(16)) = 3.125 seconds and for the height (y): D = 100(3.125) - 16(3.125)^2 = 156.25 ft. Since the max height of the ball is 156.25 ft, it can not reach 160 ft.
Answered by
2
Answer:
We calculate the maximum height of the ball, we find the vertex of the equation: ((-B / 2A),f(-B / 2A)) --> x = (-100/2(16)) = 3.125 seconds and for the height (y): D = 100(3.125) - 16(3.125)^2 = 156.25 ft. Since the max height of the ball is 156.25 ft, it can not reach 160 ft.
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