A rock is thrown downward from an unknown

height above the ground with an initial speed
of 28 m/s. It strikes the ground 4.4 s later.
Determine the initial height of the rock
above the ground. The acceleration of gravity
is 9.8 m/s
2
.
Answer in units of m
014 10.0 points
An object is released from rest on a planet
that has no atmosphere. The object falls
freely for 5.76 m in the first second.
What is the magnitude of the acceleration
due to gravity on the planet?
Answer in units of m/s

To determine the initial height of the rock above the ground, we can use the equation of motion:

h = h0 + v0t + 0.5at^2

Where:
h = final height (0 in this case, as it strikes the ground)
h0 = initial height
v0 = initial velocity (28 m/s)
t = time (4.4 s)
a = acceleration due to gravity (-9.8 m/s^2)

Plugging in the values:

0 = h0 + (28 m/s)(4.4 s) + 0.5(-9.8 m/s^2)(4.4 s)^2

Simplifying:

0 = h0 + 123.2 m - 96.416 m

Rearranging the equation to solve for h0:

h0 = -123.2 m + 96.416 m

h0 = -26.784 m

Since we can't have a negative height, the initial height of the rock above the ground is 26.784 m.

To determine the acceleration due to gravity on the planet, we can use the equation:

d = 0.5gt^2

Where:
d = distance fallen (5.76 m)
g = acceleration due to gravity on the planet (what we're trying to find)
t = time (1 second)

Plugging in the values:

5.76 m = 0.5g(1 second)^2

Simplifying:

5.76 m = 0.5g

Rearranging the equation to solve for g:

g = (5.76 m) / (0.5)

g = 11.52 m/s^2

The magnitude of the acceleration due to gravity on the planet is 11.52 m/s^2.

To determine the initial height of the rock above the ground, we can use the equation of motion for an object in free fall:

s = ut + (1/2)at^2

where s is the displacement (change in height), u is the initial velocity, t is the time, and a is the acceleration due to gravity.

Given that the rock is thrown downward with an initial velocity of 28 m/s and it takes 4.4 s to strike the ground, we can substitute these values into the equation:

0 = (28)(4.4) + (1/2)(9.8)(4.4)^2

Simplifying the equation, we have:

0 = 123.2 + 96.176

Rearranging the equation to solve for the initial height:

123.2 = 96.176

Hence, the initial height of the rock above the ground is 96.176 m.

To determine the magnitude of the acceleration due to gravity on the planet in the second problem, we can use the equation of motion again:

s = (1/2)at^2

Given that the object falls freely for 5.76 m in the first second, we can substitute these values into the equation:

5.76 = (1/2)(a)(1)^2

Simplifying the equation, we have:

5.76 = (1/2)(a)

Multiplying both sides by 2 to solve for acceleration:

11.52 = a

Hence, the magnitude of the acceleration due to gravity on the planet is 11.52 m/s^2.

Initial speed = 28m/s

time = 4.4s
Acceleration = 9.8

Just plugin the numbers into this formula and u will get the answer.

S=ut+1/2*a*t^2

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I know it breaks your heart
Moved to the city in a broke-down car
And four years, no call
Now I'm looking pretty in a hotel bar
And I, I, I, I, I can't stop
No, I, I, I, I, I can't stop

So, baby, pull me closer
In the back seat of your Rover
That I know you can't afford
Bite that tattoo on your shoulder
Pull the sheets right off the corner
Of that mattress that you stole
From your roommate back in Boulder
We ain't ever getting older

We ain't ever getting older
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So, baby, pull me closer
In the back seat of your Rover
That I know you can't afford
Bite that tattoo on your shoulder
Pull the sheets right off the corner
Of that mattress that you stole
From your roommate back in Boulder
We ain't ever getting older

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No, we ain't ever getting older