Physics, 31.03.2020 00:12 whatnowb

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(b) A student wants to know how the motion of the rock would be different if it was thrown upward at 15 m/s from a height of 100 m above Earth's surface in a clear, coherent paragraph-length
response that may also contain figures and/or equations, explain how the motion of the rock on Earth will be different from its motion on Planet Xin ems of its maximum height above the
ground, the speed at which it reaches the ground, the time in which it is in free fall, and its acceleration due to gravity

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Answers

The acceleration due to gravity is a factor of motion that determines the time, height reached, and velocity of an object moving in free fall motion

(b) The differences between the motion of the of the rock on Planet X and on Earth are;

The acceleration due gravity on Earth is almost twice on EarthThe rock spends less time in free fall on EarthThe maximum height reached by the rock on Earth is approximately half the height it gets to on Planet XThe speed of the rock as it reaches the ground on Earth is approximately 11.8 m/s faster on Earth than on Planet X

The reason the differences are correct is as follows;

Question: The part (a) part of the question appear to be as follows;

On Planet X, a rock is thrown upwards by an astronaut at a speed of 15 m/s from a height of 100 m and it takes 10 seconds to reach the ground

The information from part (a) gives;

$s = u \cdot t + \dfrac{1}{2} \cdot a\cdot t^2$

We get;

$0 = 100 + 15 \times 10 - \dfrac{1}{2} \times a\times 10^2$

$\dfrac{1}{2} \times a\times 10^2 = 250$

$a = \dfrac{250 \times 2}{10^2} = 5$

The acceleration due to gravity on Planet X, a = 5 m/s²

The velocity at ground level is found as follows;

v² = 15² + 2 × 5 × 100 = 1225

Velocity at ground level, $v = \sqrt{(1,225 \ m^2/s^2) }$ = 35 m/s

$The \ \mathbf{maximum \ height} \ on \ Planet \ X = \dfrac{(15 m/s)^2}{(2\times 5 m/s^2)}$ = 22.5 m

The acceleration due to gravity on Planet X is 5 m/s²

The motion of the rock on Earth is as follows;

On planet Earth, the acceleration due to gravity, g is approximately 9.81 m/s² which is almost twice the acceleration due to gravity on Planet X

The time the rock is in motion in the atmosphere of Earth is given as follows;

$0 = 100 + 15 \times t - \dfrac{g \times t^2}{2}$

Therefore

$0 = \dfrac{g \times t^2}{2} - 100 - 15 \times t$

Solving gives;

$(t + 3.24) \cdot (t - 6.296) = 0$

The time it takes the rock to reach the ground on Earth is approximately 6.296 seconds, therefore,

The rock spends a lesser time moving in free fall in the air on Earth than in motion on Planet X

$The \ maximum \ height \ reached \ on \ Earth = \dfrac{(15 m/s)^2}{(2\times 9.81 m/s^2) } \approx 11.5 \ m$

Therefore;

The rock gets to approximately half the height it gets to on Planet X

Speed at which it reaches the ground on Earth, v, is given as follows;

$v = \sqrt{((15 m/s)^2 + 2\times 9.81 m/s^2 \times 100 m) } \approx 46.8 m/s$

Therefore;

The rock reaches the ground, moving faster on Earth than on Planet X

Learn more about acceleration due to gravity here:

link

Answer from: millie54

Speed at which it will reach the ground is given as

v_f = 46.8 m/s

Total time for which it will remain in air is given as

t = 6.3 s

Explanation:

As we know that the object is projected upwards with speed

v_i = 15 m/s

g = - 9.81 m/s^2

now when it will reach the ground then we have

$y = v_y t + \frac{1}{2} at^2$

so we have

$-100 = 15 t - \frac{1}{2}(-9.81) t^2$

4.905 t^2 - 15 t - 100 = 0

so we have

t = 6.3 s

Now speed of the object when it reaches the ground is given as

v_f = v_i + at

v_f = -15 + (9.81)(6.3)

v_f = 46.8 m/s

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