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Arisa [49]
3 years ago
5

A punter wants to kick a football so that the football has a total flight time of 4.70s and lands 56.0m away (measured along the

ground). Neglect drag and the initial height of the football.
How long does the football need to rise?

What height will the football reach?

With what speed does the punter need to kick the football?

At what angle (θ), with the horizontal, does the punter need to kick the football?
Physics
1 answer:
Sindrei [870]3 years ago
3 0

Answer:

Explanation:

How long does the football need to rise?

4.70/3 = 2.35 s

What height will the football reach?

h = ½(9.81)2.35² = 27.1 m

With what speed does the punter need to kick the football?

vy = g•t = 9.81(2.35) = 23.1 m/s

vx = d/t = 56.0/4.70 = 11.9 m/s

v = √(vx²+vy²) = 26.0 m/s

At what angle (θ), with the horizontal, does the punter need to kick the football?

θ = arctan(vy/vx) = 62.7°

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A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.5 m>s. Two seconds later
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Answer:

4.28 s

Explanation:

after two seconds (2 s) His friends is

d = 3.5 m/s x 2 s = 7 meter ahead.

in this state, a bicylist start from initial velocity vo = 0 m/s and accelerat 2.4 m/s²

then, when bicylist reach his friend

t friend = t bicyclist = t

d bicylist = d friend + d

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d friend = 3.5 . t

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d friend + d = vo . t + ½ a t²

3.5 t + 7 = 0 . t + ½ . 2.4 . t²

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Please solve this asap​
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Let \vec u and \vec v be the vectors, and let x=\|\vec u\|=\|\vec v\| be their common magnitude.

The resultant \vec u + \vec v is \sqrt 2 times larger in magnitude than either vector alone, so \|\vec u+\vec v\| = \sqrt2\,x.

Recall the dot product identity

\vec a \cdot \vec b = \|\vec a\| \|\vec b\| \cos(\theta)

where \theta is the angle between the vectors \vec a and \vec b. In the special case of \vec a=\vec b, we get

\vec a \cdot \vec a = \|\vec a\|^2 \cos(0^\circ) \implies \|\vec a\| = \sqrt{\vec a\cdot\vec a}

Now, to get the angle between \vec u and \vec v, we have

\vec u \cdot \vec v = \|\vec u\| \|\vec v\| \cos(\theta) \implies \cos(\theta) = \dfrac{\vec u \cdot \vec v}{x^2}

To compute the dot product, we take the dot product of the resultant with itself.

(\vec u+\vec v) \cdot (\vec u + \vec v) = \|\vec u + \vec v\|^2

Solve for \vec u\cdot\vec v.

(\vec u\cdot\vec u) + 2(\vec u\cdot\vec v) + (\vec v\cdot\vec v) = \|\vec u + \vec v\|^2

\|\vec u\|^2 + 2(\vec u\cdot \vec v) + \|\vec v\|^2 = \|\vec u+\vec v\|^2

x^2 + 2(\vec u \cdot \vec v) + x^2 = (\sqrt2\,x)^2

2(\vec u\cdot\vec v) + 2x^2 = 2x^2

2(\vec u\cdot\vec v) = 0

\vec u\cdot\vec v = 0

Since their dot product is zero, \vec u and \vec v are perpendicular, so \theta=90^\circ.

7 0
2 years ago
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