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A ball is thrown vertically upward. After t seconds, its height h, in feet, is given by the function LaTeX: h\left(t\right)=-16t^2+64th ( t ) = − 16 t 2 + 64 t. How many seconds will it take the ball to reach its maximum height?

User Zmb
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2 Answers

1 vote

It will take 2 seconds for the ball to reach its maximum height.

To find the time it takes for the ball to reach its maximum height, we need to determine the time at which the velocity of the ball becomes zero. The maximum height is reached when the vertical velocity is zero.

The velocity function (v) is the derivative of the height function (h) with respect to time (t). So, we need to find the derivative of the given height function:

h(t)=−16t^2+64t

Let's find v(t)= dh/dh =−32t+64

Now, set v(t) equal to zero and solve for t:

−32t+64=0

Solving for t:

−32t=−64

t= −64/−32​ =2

So, it will take 2 seconds for the ball to reach its maximum height.

User Itsmcgh
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3 votes

Answer:

The time it will take for the ball to reach maximum height of 64 feet is 2 seconds

Explanation:

The given equation for the height, h, in feet the ball reaches is given by the function;

h(t) = -16·t² + 64·t

To reach maximum height,
h_(max), we have;

At
h_(max), we have;


\frac{\mathrm{d} h}{\mathrm{d} t} = 0


\frac{\mathrm{d} \left (-16\cdot t^(2) + 64\cdot t \right )}{\mathrm{d} t} = 0

Which gives;

64 -32·t = 0

64 = 32·t

t = 64/32 = 2 seconds

Therefore, it will take 2 seconds for the ball to reach maximum height

The maximum height reached is given as follows;

h(2) = -16×2² + 64×2 = 64 feet

The time it will take for the ball to reach maximum height of 64 feet is 2 seconds.

User Kchan
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