Answer:
The answer to your question is:
a) t = 3.81 s
b) vf = 37.4 m/s
Explanation:
Data
height = 71.3 m = 234 feet
t = 0 m/s
vf = ?
vo = 0 m/s
Formula
h = vot + 1/2gt²
vf = vo + gt
Process
a)
h = vot + 1/2gt²
71.3 = 0t + 1/2(9.81)t²
2(71.3) = 9,81t²
t² = 2(71.3)/9.81
t² = 14.53
t = 3.81 s
b)
vf = 0 + (9.81)(3.81)
vf = 37.4 m/s
<u>Answer:</u>
The height of ramp = 124.694 m
<u>Explanation:</u>
Using second equation of motion,

From the question,
u = 31 m/s; s = 156.3 m, a=0
substituting values

t = 
= 5.042 s
Similary, for the case of landing
t = 5.042 s; initial velocity, u =0
acceleration = acceleration due to gravity, g = 9.81 
Substituting in 

h = 124.694 m
So height of ramp = 124.694 m
That is true imo not false
Assume the snow is uniform, and horizontal.
Given:
coefficient of kinetic friction = 0.10 = muK
weight of sled = 48 N
weight of rider = 660 N
normal force on of sled with rider = 48+660 N = 708 N = N
Force required to maintain a uniform speed
= coefficient of kinetic friction * normal force
= muK * N
= 0.10 * 708 N
=70.8 N
Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.
Answer: 113.75
Explanation:
You know
acceleration = a = 3.5 m/s²
time = t = 5 seconds
initial velocity = u = 14 m/s
Unknown is distance = s = ?
Use equation: s = ut +
at²
Substitute all the known values inside the equation:
s = (14*5) + 0.5 * 3.5 * 5²
s = 70 + 43.75 = 113.75 m
The car travels 113.75 metres.