Answer:

Explanation:
Let the hails are coming down with speed "v" and then rebound with the same speed in opposite direction
So here the change in the momentum of the hails is given as


now we know by Newton's 2nd Law

here we have

now plug in all data


Answer:
T = 92.8 min
Explanation:
Given:
The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

Find:
- How long does the International Space Station take to orbit the earth? Give an exact answer.
Solution:
- Using the the expression given we can extract the angular speed of the International Space Station orbit:

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8
- We know that the relation between angular speed w and time period T of an orbit is related by:
T = 2*p / w
T = 2*p / (2*p / 92.8)
Hence, T = 92.8 min
Answer:
3.25 seconds
Explanation:
It is given that,
A person throws a baseball from height of 7 feet with an initial vertical velocity of 50 feet per second. The equation for his motion is as follows :

Where
s is the height in feet
For the given condition, the equation becomes:

When it hits the ground, h = 0
i.e.

It is a quadratic equation, we find the value of t,
t = 3.25 seconds and t = -0.134 s
Neglecting negative value
Hence, for 3.25 seconds the baseball is in the air before it hits the ground.
Work=applied Force x distance
= 1275 x 26
=33150 Joules
225 = 1/2 (50) (v2)
225 = 25 (v2)
225/25 = v2
9 = v2
√9 = v
v = 3 m/s