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NARA [144]
1 year ago
6

A football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees. What is the vertical component of t

he initial velocity?
Physics
1 answer:
Svetllana [295]1 year ago
4 0

If a football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees, then the vertical component of the initial velocity would be 12.65 m/s

<h3>What is Velocity?</h3>

The total displacement covered by any object per unit of time is known as velocity. It depends on the magnitude as well as the direction of the moving object. The unit of velocity is meter/second.

As given in the problem A football is kicked from the ground with a speed of 16.71 m/s at an angle of 49.21 degrees

The horizontal component of the velocity is given by

Vx = Vcosθ

The vertical component of the velocity is given by

Vy = Vsinθ

As we have to find the vertical component of the velocity given that speed of 16.71 m/s at an angle of 49.21 degrees from the ground

Vy = 16.71 × sin49.21°

Vy = 12.65 m/s

Thus, the vertical component of the velocity would be 12.65 m/s

Learn more about Velocity from here

brainly.com/question/18084516

#SPJ1

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Answer:

2

Explanation:

To find force it's force = mass times acceleration so to find mass you would divide force by acceleration

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If the ball that is thrown downward has an acceleration of magnitude aaa at the instant of its release (i.e., when there is no l
kakasveta [241]

Explanation:

At the instant of release there is no force but an acceleration of a, this means the ball is falling freely under the force of gravity. Then the acceleration would be due to force of gravity and acceleration a = g =9.81 m/s^2.

g= acceleration due to gravity

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2 years ago
Galileo Galilei was the first scientist to perform experiments in order to test his ideas. He was also the first astronomer to s
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Explanation:

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6 0
3 years ago
PE=30J, m=?, g=10m/s2, h=10m
OleMash [197]
Based on the given, this is probably a gravitational potential energy problem (PEgrav). The formula for PEgrav is:

PEgrav = mgh

Where:
m = mass (kg)
g = acceleration due to gravity
h = height (m)

With this formula you can derive the formula for your unknown, which is mass. First put in what you know and then solve for what you do not know.

PEgrav=mgh
30J=m(10)(10[tex] \frac{30}{100} =m)[/tex]

Do operations that you can with what is given first.

30J=m(100m)

Transpose the 100 to the other side of the equation. Do not forget that when you transpose, you do the opposite operation.

\frac{30}{100} =m

m = 0.30kg

5 0
3 years ago
Un objeto se suelta desde determinada altura y emplea un tiempo t en caer al suelo. Si se cuadruplica la altura desde la cual se
blondinia [14]

When an object falls from a h height, you should work with the uniformly accelerated linear movement equations:

y=½*a*t²+Vo*t+yo

You should consider:

a=-g=-10m/s²

yo=h

If it’s a freefall, it means it starts from rest, which means it has no initial velocity:

Vo=0

Replacing that information in the equation:

y=½*(-10m/s²)*t²+0*t+h=-5m/s²*t²+0+h=-5m/s²*t²+h

So this is the

Besides, if you want to find out how long it takes for it to get to the floor, you should put the height of the floor as final height, which would be 0 (assuming the initial height has been measured from there):

y=0

0=-5m/s²*t²+h

5m/s²*t²=h

t²=h/(5m/s²)

t=√(h/(5m/s²))

t=√(hs²/(5m))

t=(√(h/(5m)))s

<span>If we <span>quadruple </span>h:</span>

t2=(√(h2/(5m)))s=(√(4*h1/(5m)))s=(√4)*(√h1/(5m)))s=2*(√h1/(5m)))s=2*t1

This 4 goes inside the square root, so then it converts to 2. So the new time is twice as much the previous time.

Concerning velocity, you have to use the other equation:

v=at+vo

As I said before, a is gravity and vo is zero.

v=-10m/s²*t+0=-10m/s²*t

Final velocity is directly related to time, so if time is doubled, so is velocity.

v2=-10m/s²*t2=-10m/s²*(2*t1)=2*(-10m/s²*t1)=2*v1

<span>So the correct answer is A, and the other ones are false.</span>

8 0
3 years ago
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