- #1

dekoi

The projectile does not start from rest.

If i use the formula: vfy = viy - gt, i can't solve for v, since the value for g is unknown.

Thank you .

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- Thread starter dekoi
- Start date

- #1

dekoi

The projectile does not start from rest.

If i use the formula: vfy = viy - gt, i can't solve for v, since the value for g is unknown.

Thank you .

- #2

hotvette

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- #3

dekoi

Which equation for motion?

I still do not understand.

I still do not understand.

- #4

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dekoi said:The projectile does not start from rest.

You're sure this is correct and you've listed everything that's been given to you (v

[tex]

v_{y1} = v_ {y0} + g't \Longrightarrow g' = \frac{v_{y1}-v_{y0}}{t}

[/tex]

You could then put v

- #5

dekoi

A physics student on Planet Exidor throws a ball, and it follows a parabolic trajectory. The ball's position is shown at 1 s intervals until t = 3 s. At t = 1 s, the ball's velocity is v = (2.0i + 2.0j) m/s. (i and j being unit vectors for x and y)

a) Determine the ball's velocity at t = 0s, 2s, and 3s.

b) What is the value of g on Planet Exidor?

c) What was the ball's launch angle?

- #6

Päällikkö

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- #7

- #8

Päällikkö

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[tex]v_y = v_{y0} - gt[/tex]

What do you know at t = 1s and t = 2s?

What do you know at t = 1s and t = 2s?

- #9

dekoi

At t = 1s, vx = 2.0 m/s and vy = 2.0 m/s.

At t = 2s, vx = 2.0 m/s and vy = (2.0 - g) m/s

At t = 2s, vx = 2.0 m/s and vy = (2.0 - g) m/s

- #10

Fermat

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So the time to reach max height = 2s

You also know the vertical velocity at t = 1.

By symmetry the velocity at t = 3 will be the same as at t= 1, but in the opposite direction.

Now you can use your eqns of motion to solve for g, etc.

- #11

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dekoi said:At t = 2s, vy = (2.0 - g) m/s

{quote snipped for clarity}

This answer is correct; however, by looking at your graph I can tell you the *exact* numerical value of v

- #12

dekoi

v2y= 0 m/s

therefore

0 m/s = 2.0 - g

g = 2.0 m/s2

v0 = (2.0i + 4.0j) m/s

v1 = (2.0i + 2.0j) m/s

v2 = (2.0i) m/s

v3 = (2.0i - 2.0j) m/s

v0 = sqrt(2.0^2 + 4.0^2) = sqrt(4 + 16) = sqrt20 m/s

angle = arccos(v0x / v0) = arccos(2 / sqrt20) = 1.1 degrees

I am not sure about the angle.

- #13

dekoi

Any responses?

- #14

Päällikkö

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You have radians on.

Looks good.

Looks good.

- #15

dekoi

Thank you everyone.

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