Question

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?

Answer 1

Answer:

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

Explanation:

Short Explanation

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

• $v$ is the final velocity of the object,
• $u$ is the initial velocity of the object,
• $a$ is the acceleration (should be constant,) and
• $x$ is the displacement of the object while its velocity changed from $v$ to $u$.

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

• $v$ 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 $v$:

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

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

Explanation

Let $v$ 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 $u$ 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 $v$:

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

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