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
Answer is miles sir your welcome it was simple
Answer:
top speed = 17.25
Total height = 281.19 m
Explanation:
given data
mass = 75 kg
thrust = 160 N
coefficient of kinetic friction = 0.1
solution
we get here frictional force acting that is
frictional force =
.............1
frictional force = 0.1 × 75 × 9.8
frictional force = 73.5 N
and
Net force acting will be F = 160 - 73.5 N
F = 86.5 N
so
Acceleration in the First 15 second will be
F = ma .........2
86.5 = 75 × a
a = 1.15 m/s²
and
now After 15 second the velocity will be as
v = u + at ..........3
here u is 0
so v will be
V = 1.15 × 15
v = 17.25
and
now we get travels distance S in 15 s
s = u × t + 0.5 × a × t²
here u is 0
so distance s will be
s = 0.5 × a × t²
s = 0.5 × 1.15 × 15²
s = 129.37 m
and
now acceleration acting is
F =
m a =
a = 
a = - 0.98
here it is negative it mean downward nature of acceleration
and
now we get distance s by this formula
V² - u² = 2 a s
here v velocity is 0 and
u initial velocity is 17.25 m/s
put here value
0 - 17.25² = 2 × (-0.98) × s
solve it we get
s = 151.82 m
so
Total height is
Total height = 129.37 m + 151.82 m
Total height = 281.19 m
length of the grilling machine is 1.2 m
time taken to cook the burger is 2.7 min = 162 s
so the speed of the machine should be like this that if must have to cook till it cross the machine



now in one minute the total length of the machine that is covered is given by


now distance between the burgers is 15 cm
so total production rate will be

so it will produce 3 burger per minute