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scZoUnD [109]
3 years ago
11

An astronaut holds a rock 100m above the surface of Planet XX. The rock is then thrown upward with a speed of 15m/s, as shown in

the figure. The rock reaches the ground 10s after it is thrown. The atmosphere of Planet XX has a negligible effect on the rock when it is in free fall. Determine the acceleration due to gravity of the rock when it is on Planet XX.
Physics
1 answer:
mixer [17]3 years ago
8 0

Answer:

Explanation:

Using the equation of motion to get the acceleration due to gravoty of the rock on the planet.

S = ut+ 1/2at² where;

S is the distance of the rock above the surface of the planet = 100m

u is the initial velocity = 15m/s

a is the acceleration due to gravity

t is the time taken by the rock to reach the ground = 10s

Since the rock is thrown upward the acceleration due to gravity will be negative i.e a= -g

The equation becomes S = ut- 1/2gt²

Substituting the given value to get the time t

100 = 15(10)- 1/2g(10)²

100 = 150-50g

100-150 = -50g

-50 = -50g

g = -50/-50

g = 1m/s²

<em>Hence the acceleration due to gravity of the rock when it is on Planet XX is 1m/s²</em>

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Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

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