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Molodets [167]
3 years ago
7

Chris shoots an arrow up into the air. The height of the arrow is given by the function h(t) = - 16t2 + 64t + 23 where t is the

time in seconds. What is the maximum height of the arrow?
Physics
2 answers:
Ghella [55]3 years ago
7 0

Completing the square gives the answer right away.

-16t^2+64t+23=-16(t^2-4t)+23=-16(t^2-4t+4-4)+23=-16((t-2)^2-4)+23

\implies h(t)=-16(t-2)^2+87

which indicates a maximum height of 87 when t=2.

vagabundo [1.1K]3 years ago
6 0

Answer:

The maximum height of the arrow is 87 meters.

Explanation:

If we look at the height function of the arrow

                h(t)=-16t^{2} +64t+23

we see that its a parabola whose principal coefficient is negative, that means is inverted or upside down.

When the arrow reaches maximum height its velocity will be zero. The velocity of an object is the derivative of the position function, in this case the so called height function.

So we we derivate the height function to get that

                h'(t)=-32t+64

we must find the t that makes this equation equal to zero:

                -32t+64=0

                          32t=64

                             t=2s

we replace this value of t in the height function:

                h(2 s)=-16.(2s)^{2} +64.(2s)+23

we get that

                 h(2s)=87m

The maximum height of the arrow is 87 meters.

We have used the MKS system which uses the meter, kilogram and second as base units.

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A) 140 degrees

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Assuming the wheel is moving at constant angular velocity, we can now calculate the angular displacement with respect to the initial position:

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and substituting t = 75 seconds, we find

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In degrees, it is

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The tangential speed, v, of a point at the egde of the wheel is given by

v=\omega r

where we have

\omega=0.20 rad/s

r = d/2 = (27 m)/2=13.5 m is the radius of the wheel

Substituting into the equation, we find

v=(0.20 rad/s)(13.5 m)=2.7 m/s

6 0
3 years ago
While playing baseball with your friends your hands begin to sting after you ctach several fast balls.
muminat

Answer:

Explanation:

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eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

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X_f = \frac{1}{2} (V_i+V_f)t

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Hope this helps! :)
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