Explanation:
Given that,
Initial speed of the rock, u = 30 m/s
The acceleration due to gravity at the surface of the moon is 1.62 m/s².
We need to find the time when the rock is ascending at a height of 180 m.
The rock is projected from the surface of the moon. The equation of motion in this case is given by :

It is a quadratic equation, after solving whose solution is given by:
t = 7.53 s
or
t = 8 seconds
(e)If it is decending, v = -20 m/s
Now t' is the time of descending. So,

Let h' is the height of the rock at this time. So,

or
h' = 155 m
Answer:
Some stars in space can come to life causing humans to be amazed but at thesame time in great/critical danger.
Answer:
V = 0.9 m/s
Explanation:
The parameters given are:
Initial velocity U = 6.4 m/s
Time t = 0.64s
Height h = 2.05 m
To find the final velocity, let us use third equation of motion
V^2 = U^2 - 2gH
Since the ball is going upward, g will be negative
Substitute all the parameters into the formula
V^2 = 6.4^2 - 2 × 9.8 × 2.05
V^2 = 40.96 - 40.18
V^2 = 0.78
V = sqrt( 0.78)
V = 0.883 m/ s
V = 0.9 m/ s approximately