A basketball player throws a ball vertically with initial velocity of 100ft/sec . the distance of the ball from the ground after 1 second is given by the expression 100t - 16t²

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 50ft. from the ground?

3) how many second will it take for the ball to fall from the ground ?

4) do you think the ball can reach the height of 160 ft ?. why or why not ?

can you please answer this ? i really need your help guysss ....

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Answers

2014-08-08T18:04:13+08:00
1. Since we have the formula, we substitute 6 s directly to the equation: D = 100(6) - 16(6)^2 = 24 ft. 2. We are given the distance, so we write: 50 = 100T - 16T^2 --> 2(8T^2 - 50T - 25) = 0 --> T = -3.6 and 3.6. Since we can't use a negative time, we use 3.6 seconds. 3. 0 = 100T - 16T^2 --> 0 = T (100 - 16T) --> T = 0, 6.25. Since 0 seconds can not be possible, we use 6.25 seconds. 4. 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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