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2 years ago

From what height should you drop a ball so that it hits the ground at 40 m/s? How long will it take to hit the ground?

Physics

1 answer:

Ksenya-84 [330]2 years ago

7 0

Explanation:

It is given that,

A ball hits the ground at 40 m/s

Let h is the height from where you drop a ball. It is based on the conservation of energy as :

$mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}\\\\\text{Putting all the values in above formula}\\\\h=\dfrac{40^2}{2\times 9.8}\\\\h=81.63\ m$

It is dropped from a height of 81.63 m.

Let it take t seconds to hit the ground. When it hits the ground, its final speed, v = 0 and u = 40 m/s. So,

$h=ut+\dfrac{1}{2}gt^2\\\\h=\dfrac{1}{2}gt^2\\\\t=(\dfrac{2h}{g})^{1/2}\\\\t=(\dfrac{2\times 81.63}{9.8})^{1/2}\\\\t=4.08\ s$

So, it will take 4.08 seconds to hit the ground.

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Answer:

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2 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

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$\theta=\dfrac12\alpha t^2$

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$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2$

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

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d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

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which would be equal to

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3 years ago

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Slav-nsk [51]

Answer:

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Answer:

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Explanation:

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