Olivia is cutting a 1 \dfrac12 \text{ m}1 2 1 m1, start fraction, 1, divided by, 2, end fraction, start text, space, m, end te
2 answers:
You might be interested in
Answer:
Step-by-step explanation:
Given that,
The function f(t) = 3t − 0.5t² is the height of the motion of the stone in 10m units
And it's initial velocity is 30m/s
u=30m/s
Using calculation
Let find it turning point
f(t) = 3t − 0.5t²
Then, f'(t) =3-t
f(t)=0
So, 0=3-t
Then t=3secs
So let know the inflection point to show if t=3secs is the maximum point or minimum point
Then, we need second integral of f(t)
f'(t) = 3 − t
f''(t)=-1
Since f''(t) is negative this shows that the point is the maximum point
So at t=3secs is the maximum point,
So now to know the maximum point
f(t) = 3t − 0.5t²
t=3secs
f(t)=3(3)-0.5(3)²
f(t)=9-4.5
f(t)=4.5m
Since f(t) is in 10m units
Then,
The height is f(t) =10×4.5
f(t) =45m
Then the maximum height of the stone is 45m and the time to reach the maximum height is t=3secs
Now, Using equation of motion
Time to reach max height.
v=u-gt. Upward motion g=9.81m/s2
Final velocity is v=0m/s
0=30-9.81t
-30=-9.81t
t=-30/-9.81
t=3.06secs
t=3.1secs. To 1d.p
If we have used g=10m/s², it will have the same value as the graph and the maximum and minimum point calculation
Maximum height, using equation of motion
v²=u²-2gH
0²=30²-2×9.81H
-30²=-19.62H
Then,
H=-30²/-19.62
H=45.87
a. From the graph it shows that the maximum value of the is 4.5m at time t=3secs
b. Now, the maximum height of stone as shown on the graph is 4.5m
And since is 10 meter unit scale
Then the height becomes 4.5×10
Maximum height is 45m.
c. The stone will spend two times the time it uses to reach t maximum height to return to the height of the cliff
Time to return back to the cliff is
T=2t
T=2×3
T=6secs
d. If the stone hits the ground after t=7secs
So after the ball reached a maximum height of 45m, he has spent 3secs, so we can calculate the distance the stone travelled from the maximum to the ground
The time is of travel will be 4secs
Then,
Initial velocity is 0, from the point of return at rhe maximum height
S=ut+½gt²
S=0+½×9.81×4²
S=78.48m
So the total distance from the maximum height down to the bottom of the cliff is 78.48m
Then, the height of the cliff is the height from the maximum height to the bottom of the cliff minus the maximum height reach by the stone
Then,
H(cliff)=78.48-45.87
H(cliff)=33.48m
From the graph
At t=7secs
The stone is at a distance of -3.5m
Showing that the height of the cliff
Since it is of 10m units
Then the height of the cliff is
3.5×10=35m
Also
f(t) = 3t − 0.5t² at t=7secs
F(7)=3(7)-0.5(7)²
F(7)=21-0.5×49
F(7)=21-24.5
Then,
f(7)=-3.5m
This shows the downward motion
Since it is 10m unit
The the height of cliff downward is
3.5×10=35m
Height of cliff =35m
