Sitting at the edge of a cliff, you throw a stone vertically upward with a velocity of 30 m/s. The height of the stone above the cliff's edge in units of 10 meters can be approximated by

f(t) = 3t − 0.5t2.
(a) From the graph, at what time does the stone reach its highest point? (Round your answer to one decimal place.)
t = s
(b) How high above the cliff does the stone reach? (Round your answer to one decimal place.)
m
(c) When is the stone again at the elevation of the cliff's edge? (Round your answer to one decimal place.)
t = s
(d) You check your watch and observe that the stone hits the ground below the cliff after 7 seconds. How high is the cliff? (Round your answer to one decimal place.)

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

constant acceleration- a

constant velocity- c

no velocity or acceleration- b

explanation:

for body moving with a constant acceleration, the rate of change of displacement changes over time. this is proven by the increasing gradient of the curve (a) against time.

a body that is moving at constant velocity covers equal distances over equal periods of time, thus the straight line with a constant gradient c.

a stationary body   covers no displacement as time elapses. thus the flat line with   zero gradient.

answer: 20

step-by-step explanation: