Question

The following situation will be used for the next three problems: A rock is projected upward from the surface of the moon, at time t = -0.0s, with a velocity of 30m/s. The acceleration due to gravity at the surface of the moon is 1.62m/s2 the time when the rock is ascending at a height of 180m is closest to:______.
a. 8s .
b. 12s.
c. 17s.
d. 23s.
e. 30s
For the previous situation, the height of the rock when it is descending with a velocity of 20m/s is closest to:_____.
A. 115m.
B. 125m.
C. 135m.
D. 145m
E. 155m.

Answer 1

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 :

$h=ut-\dfrac{1}{2}gt^2\\\\180=30t-\dfrac{1}{2}\times 1.62t^2$

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,

$v=-u+gt\\\\t=\dfrac{v+u}{g}\\\\t=\dfrac{20+30}{1.62}\\\\t=30.86\ s$

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

$h'=ut-\dfrac{1}{2}gt^2\\\\h'=30\times 30.86-\dfrac{1}{2}\times 1.62\times 30.86^2\\\\h'=154.40\ m$

or

h' = 155 m