The height of a basketball $t$ $t$ seconds after it is thrown can be modeled by the function $y = - 16 t^{2} + 32 t + 2$ $y=-16t^2 + 32t + 2$ . What is maximum height of the basketball? How long is the basketball in the air, if its caught in its descent 7 feet above the ground?

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The height of a basketball $t$ $t$ seconds after it is thrown can be modeled by the function $y = - 16 t^{2} + 32 t + 2$ $y=-16t^2 + 32t + 2$ . What is maximum height of the basketball? How long is the basketball in the air, if its caught in its descent 7 feet above the ground?

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Answer 1 · 2017-04-14T04:44:28

Maximum height is $y = 18 f t$ $y=18 ft$ and time in air is $t ≅ 1.83 sec$ $t~=1.83 sec$

Explanation:

graph{-16x^2 + 32x + 2 [-20.67, 23.72, -1.34, 20.86]}
Consider this graph where height (y) is on y axis and time (t) is on x axis
You can imagine that at maximum height basketball's velocity becomes zero ,so it sort of hovers in air and does not go up or down
for a very short time.

So at maximum height slope of $y \Rightarrow t$ $y=>t$ graph becomes zero

For given equation differentiating with respect to time we get velocty

$y = - 16 t^{2} + 32 t + 2$ $y=-16t^2 + 32t + 2$

$\frac{\partial y}{\partial t} = - 32 t + 32$ $(dely)/(delt)=-32t+32$

$\frac{\partial y}{\partial t}$ $(dely)/(delt)$ is velocity which at maximum height is zero

$\frac{\partial y}{\partial t} = - 32 t + 32 = 0$ $(dely)/(delt)=-32t+32 =0$

$- 32 t = - 32$ $-32t=-32$

$t = 1$ $t=1$

Hence at $t = 1$ $t=1$ second ball is at maximum height which makes sense if you look at graph given above,at $t = 1$ $t=1$ second there is no change in height so it is hovering

Max height(H) $= - 16 {(1)}^{2} + 32 (1) + 2$ $=-16(1)^2+32(1)+2$ $= 18$ $=18$

It takes 1 second for the ball to reach maximum height and then it starts its descent

If we plug $y = 7$ $y=7$ in our equation we should expect two values of $t$ $t$ because ball is at $y = 7$ $y=7$ two times namely ascent and descent but we want its time during its descent hence we will take the time of larger magnitude as our answer

$7 = - 16 t^{2} + 32 t + 2$ $7=-16t^2+32t+2$

$0 = - 16 + 32 t - 5$ $0=-16+32t-5$

We get $t_{1} = 1 - \frac{\sqrt{11}}{4}$ $t_1=1 - sqrt(11)/4$ and $t_{2} = 1 + \frac{\sqrt{11}}{4}$ $t_2=1 + sqrt(11)/4$

But we will take $t_{2}$ $t_2$ as our answer

$t_{2} ≅ 1.83$ $t_2~=1.83$

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The height of a basketball $t$ $t$ seconds after it is thrown can be modeled by the function $y = - 16 t^{2} + 32 t + 2$ $y=-16t^2 + 32t + 2$ . What is maximum height of the basketball? How long is the basketball in the air, if its caught in its descent 7 feet above the ground?

1 Answer

Explanation:

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Science

Math

Humanities

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The height of a basketball tt seconds after it is thrown can be modeled by the function y=−16t2+32t+2y=-16t^2 + 32t + 2. What is maximum height of the basketball? How long is the basketball in the air, if its caught in its descent 7 feet above the ground?

1 Answer

Explanation:

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The height of a basketball $t$ $t$ seconds after it is thrown can be modeled by the function $y = - 16 t^{2} + 32 t + 2$ $y=-16t^2 + 32t + 2$ . What is maximum height of the basketball? How long is the basketball in the air, if its caught in its descent 7 feet above the ground?