Answer:
a) 6.95 m/s
b) 1.42 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²

a) The vertical speed when it leaves the ground. is 6.95 m/s

Time taken to reach the maximum height is 0.71 seconds

Time taken to reach the ground from the maximum height is 0.71 seconds
b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds
Please state the options and I will answer to the best of my abilities XD
This question involves the concepts of orbital velocity and orbital radius.
The orbital velocity of ISS must be "7660.25 m/s".
The orbital velocity of the ISS can be given by the following formula:

where,
v = orbital velocity = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of Earth = 5.97 x 10²⁴ kg
R = orbital radius = radius of earth + altitude = 63.78 x 10⁵ m + 4.08 x 10⁵ m
R = 67.86 x 10⁵ m
Therefore,

<u>v = 7660.25 m/s</u>
Learn more about orbital velocity here:
brainly.com/question/541239
Answer:
A) the maximum acceleration the boulder can have and still get out of the quarry
B) how long does it take to be lifted out at maximum acceleration if it started from rest
Explanation:
A)
let +y is upward. look below at the free body diagram. the mass M refers to the combined mass of the boulder and chain.
the weight of the chain is:
and maximum tension is 
total mass and weight is :


∑



B)
maximum acceleration

using 
to solve for t


Answer:
Explanation:
Given
Both cars mass is m
and solving problem in Vertical and horizontal direction
considering + y and +x to be positive and u be the final velocity of system
Conserving Momentum in Vertical direction

------1
Conserving momentum in x direction
-----2
squaring and adding 1 &2



