The range of the piece of paper is C) 1.4 m
Explanation:
The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:
- A uniform motion along the horizontal direction, with constant velocity
- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity,
)
From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

where
u is the initial velocity
is the angle of projection
g is the acceleration of gravity
For the piece of paper in this problem,
u = 4.3 m/s

Substituting,

Learn more about projectile motion:
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<span>This problem is relatively simple, in order to solve this problem the only formula you need to know is the formula for friction, which is:
Ff = UsN
where Us is the coefficient of static friction and N is the normal force.
In order to get the crate moving you must first apply enough force to overcome the static friction:
Fapplied = Ff
Since Fapplied = 43 Newtons:
Fapplied = Ff = 43 = UsN
and it was given that Us = 0.11, so all you have to do is isolate N by dividing both sides by 0.11
43/0.11 = N = 390.9 which is approximately 391 or C. 3.9x10^2</span>
Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s