Answer:
meters
Explanation:The question ask for the maximum value of the function f(t) which can be find by find the maxima of the function
The maxima of the function occurs when the slope is zero. i.e.

Hence the maxima occurs at t=1.63 seconds
The maximum value of f is

hence maximum height is found to be
meters
Answer:
39.7 m
Explanation:
First, we conside only the last second of fall of the body. We can apply the following suvat equation:

where, taking downward as positive direction:
s = 23 m is the displacement of the body
t = 1 s is the time interval considered
is the acceleration
u is the velocity of the body at the beginning of that second
Solving for u, we find:

Now we can call this velocity that we found v,
v = 18 m/s
And we can now consider the first part of the fall, where we can apply the following suvat equation:

where
v = 18 m/s
u = 0 (the body falls from rest)
s' is the displacement of the body before the last second
Solving for s',

Therefore, the total heigth of the building is the sum of s and s':
h = s + s' = 23 m + 16.7 m = 39.7 m
Answer:
0.4
Explanation:
F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.
Since the mass is moving with a constant velocity, there's no acceleration hence
where N is the weight of object and \mu is coefficient of kinetic friction.
the subject

Substituting F for 8 N and N for 20 N

Therefore, coefficient of kinetic friction is 0.4
Answer:

Explanation:
From the question we are told that:
Mass 
Angle 
Coefficient of static friction
Generally, the equation for Newtons second Law is mathematically given by
For


for


Where



Therefore


