Question

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?

Answer 1

Answer:

$v_f=\dfrac{g}{6}\times t_b\ m/s$

1.7985 m/s

0.989175 m

5.93505 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

$v_f=u+at\\\Rightarrow v_f=0+\dfrac{g}{6}\times t_b\\\Rightarrow v_f=\dfrac{g}{6}\times t_b\ m/s$

The expression is $v_f=\dfrac{g}{6}\times t_b\ m/s$

$v_f=\dfrac{9.81}{6}\times 1.1\\\Rightarrow v_f=1.7985\ m/s$

The final velocity is 1.7985 m/s

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow d=0\times t+\dfrac{1}{2}\times \dfrac{9.81}{6}\times 1.1^2\\\Rightarrow d=0.989175\ m$

The distance dropped from the Moon is 0.989175 m

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow d=0\times t+\dfrac{1}{2}\times 9.81\times 1.1^2\\\Rightarrow s=5.93505\ m$

The distance dropped from the Earth is 5.93505 m