Is that a question? If it is not what its the question?
Answer:
<h3>1.03684m</h3>
Explanation:
Using the formula for calculating range expressed as;
R = U√2H/g where
R is the distance moves in horizontal direction = 18.4m
H is the height
U is the velocity of the baseball = 40m/s
g is the acceleration due to gravity = 9.8m/s²
Substitute the given parameters into the formula and calculate H as shown;
18.4 = 40√2H/9.8
18.4/40 = √2H/9.8
0.46 = √2H/9.8
square both sides;
(0.46)² = (√2H/9.8)²
0.2116 = 2H/9.8
2H = 9.8*0.2116
2H = 2.07368
H = 2.07368/2
H = 1.03684m
Hence the ball is 1.03684m below the launch height when it reached home plate.
What is the question? I think that you answered it yourself...
Answer:
a)
& 
b) 
c) 
Explanation:
Given:
mass of the book, 
combined mass of the student and the skateboard, 
initial velocity of the book, 
angle of projection of the book from the horizontal, 
a)
velocity of the student before throwing the book:
Since the student is initially at rest and no net force acts on the student so it remains in rest according to the Newton's first law of motion.

where:
initial velocity of the student
velocity of the student after throwing the book:
Since the student applies a force on the book while throwing it and the student standing on the skate will an elastic collision like situation on throwing the book.

where:
final velcotiy of the student after throwing the book
b)



c)
Since there is no movement of the student in the vertical direction, so the total momentum transfer to the earth will be equal to the momentum of the book in vertical direction.


