Physics, 03.09.2020 21:01, joha222680

Gwen releases a rock at rest from the top of a 40-m tower. If g = 9.8 m/s2 and air resistance is negligible, what is the speed of the rock as it hits the ground?

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answered: kaylamorris05

$28\; \rm m \cdot s^{-1}$ .

Explanation:

Short Explanation

Apply the SUVAT equation $\left(v^2 - u^2) = 2\, a \, x$ , where:

is the final velocity of the object,

is the initial velocity of the object,

is the acceleration (should be constant,) and

is the displacement of the object while its velocity changed from

Assume that going downwards corresponds to a positive displacement. For this question:

needs to be found. u = 0

because the rock is released from rest. $a = g = 9.8 \; \rm m\cdot s^{-2}$ . $x = 40\; \rm m$ .

Solve this equation for :

$\displaystyle v = \sqrt{2\, a\, x + u^2} = \sqrt{2\times 9.8 \times 40} = 28\; \rm m \cdot s^{-1}$ .

In other words, the rock reached a velocity of $28\; \rm m\cdot s^{-1}$ (downwards) right before it hits the ground.

Explanation

Let be the velocity (in $\rm m \cdot s^{-1}$ ) of this rock right before it hits the ground. Under the assumptions of this question, it would take a time of t = (v / 9.8) seconds for this rock to reach that velocity if it started from rest and accelerated at $9.8\; \rm m \cdot s^{-2}$ .

Note that under these assumptions, the acceleration of this rock is constant. Therefore, the average velocity of this rock would be exactly one-half the sum of the initial and final velocity. In other words, if denotes the initial velocity of this rock, the average velocity of this rock during the fall would be:

$\displaystyle \frac{u + v}{2}$ .

On the other hand, u = 0 because this stone is released from rest. Therefore, the average velocity of this rock during the fall would be exactly (v / 2) .

The displacement of an object over a period of time is equal to the length of that period times the average velocity over that period. For this rock, the length of this fall would be t = (v / 9.8) , while the average velocity over that period would be (v / 2) . Therefore, the displacement (in meters) of the rock during the entire fall would be:

$\displaystyle \left(\frac{v}{2}\right) \cdot \left(\frac{v}{9.8}\right) = \frac{v^2}{19.6}$ .

That displacement should be equal to the change in the height of the rock, $40\; \rm m$ :

$\displaystyle \frac{v^2}{19.6} = 40$ .

Solve for :

$v = 28\; \rm m \cdot s^{-1}$ .

Once again, the speed of the rock would be $28\;\rm m \cdot s^{-1}$ right before it hits the ground.

answered: Guest

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answered: Guest

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