This is a defective, misleading question, and should never be asked in a Physics class.
There is no such thing as the force due to the impact.
If you know how long it takes the clam to stop once it begins to hit the dirt,
then you can calculate the impulse transferred to it, and tease a force out
of that. But the question doesn't give us the time.
It depends on the material of the surface. Was the clam dropped onto dirt ?
Into a dumpster ? Onto grass ? Concrete ? Styrofoam ? Mud ? The answer
is different in each case, and we still need to know the short length of time
AFTER it first encountered whatever surface brought it to rest.
I would kick this question back to the Physics teacher. It's meaningless,
and the longer you try to work on it, the more nonsense you'll plant into
your head that'll need to be dug out later.
Answer: i think it’s plant and glass
Explanation:
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.
Part (a): Velocity of the snowball
By conservation of momentu;
m1v1 + m2v2 = m3v3,
Where, m1 = mass of snowball, v1, velocity of snowball, m2 = mass of the hat, v2 = velocity of the hat, m3 = mass of snowball and the hat, v3 = velocity of snowball and the hut.
v2 = 0, and therefore,
85*v1 + 0 = 220*8 => v1 = 220*8/85 = 20.71 m/s
Part (b): Horizontal range
x = v3*t
But,
y = vy -1/2gt^2, but y = -1.5 m (moving down), vy =0 (no vertical velocity), g = 9.81 m/s^2
Substituting;
-1.5 = 0 - 1/2*9.81*t^2
1.5 = 4.905*t^2
t = Sqrt (1.5/4.905) = 0.553 seconds
Then,
x = 8*0.553 = 4.424 m
