Yes you're correct. For distance, SI is based on meters.
Answer:
The take-off speed is 41.48 
Explanation:
Given :
Range
m
Projectile angle
13°
From the formula of range,

Find the velocity from above equation,

( ∵
)

Therefore, the take-off speed is 41.48
The vibration is thermal energy ("heat" energy which every object possesses).
The second one is kinetic energy ("motion" energy of a massive object)
|Momentum| = (mass) x (speed)
225 kg-m/s =(50kg) x (speed)
Divide each side by (50kg): Speed=(225 kg-m/s) / (50 kg) = 4.5 m/s .
Regarding the velocity, nothing can be said other than the speed, because
we have no information regarding the direction of the object's motion.
By Newton's second law, the net vertical force acting on the object is 0, so that
<em>n</em> - <em>w</em> = 0
where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that
80 N = <em>µ</em> (196 N) → <em>µ</em> = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is
40 N = <em>ν</em> (196 N) → <em>ν</em> = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: <em>µ</em> and <em>ν</em> are the Greek letters mu and nu)