Questions: Gravitational acceleration on the moon is approximately one-sixth of that on Earth. A ball released from rest above the surface falls from height d in a time of t=12 seconds. Part (a) Write an expression for the final velocity v 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, 8 (the gravitational acceleration on Earth), and elapsed time, t. Part (b) Calculate the final velocity, yy , in meters per second. Part (c) Calculate the height d, in meters, from which the ball was dropped. Part (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?

Part (a): Write an expression for the final velocity \(v_y\) 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, \(g\) (the gravitational acceleration on Earth), and elapsed time, \(t\). Equation of motion The acceleration due to gravity on the moon is \(g_m = \frac{1}{6}g\). Since the ball starts from rest, its initial velocity \(v_0\) is 0. The final velocity can be found using the equation \(v_y = v_0 + at\), where \(a\) is the acceleration and \(t\) is the time. Substitute values Substituting \(v_0=0\) and \(a = \frac{1}{6}g\), we get \(v_y = \frac{1}{6}gt\). \(\boxed{v_y = \frac{1}{6}gt}\)

Part (b): Calculate the final velocity, \(v_y\), in meters per second. Given values \(g = 9.8 \, m/s^2\), \(t = 1.2 \, s\) Substitute values \(v_y = \frac{1}{6} (9.8 \, m/s^2) (1.2 \, s)\) Calculate \(v_y = 1.96 \, m/s\) \(\boxed{1.96 \, m/s}\)

Part (c): Calculate the height \(d\), in meters, from which the ball was dropped. Equation of motion We can use the equation \(d = v_0 t + \frac{1}{2}at^2\), where \(d\) is the distance, \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time. Substitute values Since \(v_0 = 0\) and \(a = g_m = \frac{1}{6}g\), we have \(d = \frac{1}{2} (\frac{1}{6}g) t^2 = \frac{1}{12}gt^2\). Calculate \(d = \frac{1}{12}(9.8 \, m/s^2) (1.2 \, s)^2 = 1.176 \, m\) \(\boxed{1.176 \, m}\)

\(v_y = \frac{1}{6}gt\) \(v_y = 1.96 \, m/s\) \(d = 1.176 \, m\)