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3 years ago

a kid jumping on a trampoline reaches a height of 0.925m. what was his speed when he left the trampoline?

1 answer:

3 0

As the kid rises to a height of 0.925 meters, his velocity decreases at the rate of 9.8 m/s each second. At the maximum height, his velocity is 0 m/s. Use the following equation to solve this problem.

vf^2 = vi^2 + 2 * a * d, vf = 0, a = -9.8
0 = vi^2 + 2 * -9.8 * 0.925
vi = √18.13
This is approximately 4.26 m/s.

I hope this helps, have a great day, and God bless.
Brainliest is always appreciated :)

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Answer:

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2 years ago

The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

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Part a)

At t = 0 the position of the object is given as

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At t = 2

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so displacement of the object is given as

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so average speed is given as

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

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at t = 1

$v = 2 cos(\pi/4) * \frac{\pi}{4}$

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at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

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Part d)

Now for instantaneous acceleration

As we know that

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<em>so above is the instantaneous accelerations</em>

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I think the answer is b. easy to digest.

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3 years ago

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Answer:

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$r=\frac{mv}{qB}$ (1)

If the speed of the particle is for time as fast, that is, v' = 4v, you obtain, in the equation (1):

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