Gravitational acceleration on the moon is one sixth of that on Earth. A ball released from rest above the surface falls from height d in a time of t = 1.1 seconds.a) write an expression for the final velocity vf of the ball when it impacts the surface of the moon assuming it is dropped from rest. This expression should be in terms of only, g (gravitational acceleration on Earth) and time, tb) calculate the final velocity, vf, numerically in m/sc) calculate the height, d (in meters) from which the ball was droppedd) from how high up would this ball need to be dropped on the earth, De (in meters), if it took the same time to reach the ground as it did on the moon?

Let's break the problem into parts and solve each one systematically:

### Part a) Expression for final velocity \(v_f\) on the Moon
The ball is dropped from rest, meaning the initial velocity \(v_0 = 0\). We know the formula for the final velocity \(v_f\) of an object under constant acceleration is:

\[
v_f = v_0 + at
\]

Since \(v_0 = 0\) and the acceleration on the Moon is \(\frac{g}{6}\) (where \(g\) is the gravitational acceleration on Earth, approximately \(9.8 \, \text{m/s}^2\)), the expression for the final velocity \(v_f\) on the Moon is:

\[
v_f = \left(\frac{g}{6}\right)t
\]

This is the general expression for the final velocity \(v_f\) on the Moon.

### Part b) Numerical calculation of final velocity \(v_f\)
We can now calculate the final velocity \(v_f\) using the given time \(t = 1.1 \, \text{seconds}\) and \(g = 9.8 \, \text{m/s}^2\):

\[
v_f = \left(\frac{9.8}{6}\right)(1.1)
\]

Let me calculate this.The final velocity \(v_f\) of the ball when it impacts the surface of the Moon is approximately \(1.80 \, \text{m/s}\).

### Part c) Calculation of the height \(d\) from which the ball was dropped
Using the kinematic equation for distance under constant acceleration, starting from rest:

\[
d = v_0 t + \frac{1}{2} a t^2
\]

Since \(v_0 = 0\) and the acceleration \(a = \frac{g}{6}\), the equation simplifies to:

\[
d = \frac{1}{2} \left(\frac{g}{6}\right) t^2
\]

We can substitute the known values to calculate \(d\). Let's compute that next.The ball was dropped from a height of approximately \(0.99 \, \text{meters}\) on the Moon.

### Part d) Height on Earth for the same time
On Earth, we use the same kinematic equation, but with Earth's gravitational acceleration \(g\) instead of \(\frac{g}{6}\):

\[
d_e = \frac{1}{2} g t^2
\]

Let me calculate the height \(d_e

Gravitational acceleration on the moon is one sixth of that on Earth. A ball released from rest above the surface falls from height d in a time of t = 1.1 seconds. a) Write an expression for the final velocity vf of the ball when it impacts the surface of the moon assuming it is dropped from rest. This expression should be in terms of only g (gravitational acceleration on Earth) and time, t. b) Calculate the final velocity, vf, numerically in m/s. c) Calculate the height, d (in meters) from which the ball was dropped. d) From how high up would this ball need to be dropped on the Earth, De (in meters), if it took the same time to reach the ground as it did on the moon?

Explore how to calculate the final velocity of a ball dropped from rest on the moon, using the expression vf = (g/6) * t. Additionally, learn how to find the height from which the ball was dropped and compare it to the equivalent height on Earth. Suitable for Grades 9-12.

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