A baseball on the Moon and an identical baseball on the Earth are thrown vertically upward with the same initial velocity of 12 m/s. The acceleration due to gravity on the Moon is 1/6 that of Earth. What is the maximum height the ball reaches on the moon? What is the maximum height the ball reaches on the earth? What is the difference in the maximum height the ball reaches on the Moon compared to on Earth?

a. 3.68 m b. 0.61 m c. 3.07 m

Explanation:

a. What is the maximum height the ball reaches on the moon?

Using v² = u² - 2g'h where u = initial velocity of baseball = 12 m/s, v =  final velocity of baseball = 0 m/s (since it stops at maximum height), g' = acceleration due to gravity on moon = g/6 where g = acceleration due to gravity on earth = 9.8 m/s², so g' = 9.8 m/s²/6 = 1.63 m/s² and h = maximum height of ball on moon.

So, making h subject of the formula, we have

h = -(v² - u²)/2g'

substituting the values of the variables, we have

h = -((0 m/s)² - (12 m/s)²)/2(1.63 m/s²)

=  -(- 12 m²/s²)/2(1.63 m/s²)

= 6 m²/s²)/(1.63 m/s²)

= 3.68 m

b. What is the maximum height the ball reaches on the earth?

Using the same equation for the maximum height the baseball travels on the moon for that on the earth, the maximum height h' the baseball travels on the earth is given by  

h' = -(v² - u²)/2g where u = initial velocity of baseball = 12 m/s, v =  final velocity of baseball = 0 m/s (since it stops at maximum height),  g = acceleration due to gravity on earth = 9.8 m/s²and h' = maximum height of ball on moon.

substituting the values of the variables, we have

h = -((0 m/s)² - (12 m/s)²)/2(9.8 m/s²)

=  -(- 12 m²/s²)/2(9.8 m/s²)

= 6 m²/s²)/(9.8 m/s²)

= 0.61 m

c. What is the difference in the maximum height the ball reaches on the Moon compared to on Earth?

The difference in the maximum height the ball reaches on the Moon compared to on Earth is d = h - h' = 3.68 m - 0.61 m = 3.07 m

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards