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An object falls a distance H in 50 s when dropped on the surface of the earth. How long would it take for the same object to fall through the same distance on the surface of a planet whose mass and radius are twice that of the earth ? (Neglect air resistance.)
  • a)
    35.4 g
  • b)
    50.0 s
  • c)
    70.7 s
  • d)
    100.0 s
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
An object falls a distance H in 50 s when dropped on the surface of th...
To solve this problem, we need to use the concept of gravitational acceleration and the equation for free-fall motion.

Gravitational acceleration is the acceleration experienced by an object due to the gravitational force exerted by another object. On the surface of the Earth, the gravitational acceleration is approximately 9.8 m/s^2.

The equation for free-fall motion is:
H = (1/2)gt^2,
where H is the distance fallen, g is the gravitational acceleration, and t is the time taken.

Let's solve the problem step by step:

1. Given that the object falls a distance H in 50 s on the surface of the Earth.
H = (1/2)gt^2
H = (1/2)(9.8 m/s^2)(50 s)^2
H = 12250 m

2. We need to find the time taken for the same object to fall through the same distance on the surface of a planet whose mass and radius are twice that of the Earth.

Since the mass and radius of the planet are twice that of the Earth, the gravitational acceleration on the surface of the planet will also be twice that of the Earth.

Let's assume the time taken on the planet is T.

H = (1/2)g'T^2,
where g' is the gravitational acceleration on the planet.

Using the relationship between the gravitational accelerations on the Earth and the planet:
g' = 2g,

we can rewrite the equation as:
H = (1/2)(2g)(T^2)
H = 2(g)(1/2)(T^2)
H = gT^2

Now, we can equate the two equations for H:
gT^2 = (1/2)gt^2

Canceling the g term on both sides and rearranging the equation:
T^2 = (1/2)t^2

Taking the square root of both sides:
T = √[(1/2)t^2]

Plugging in the given value of t (50 s):
T = √[(1/2)(50 s)^2]
T = √[(1/2)(2500 s^2)]
T = √[1250 s^2]
T = 35.355 s (rounded to three decimal places)

Therefore, it would take approximately 35.4 seconds for the same object to fall through the same distance on the surface of a planet whose mass and radius are twice that of the Earth.

Hence, the correct answer is option C) 35.4 s.
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An object falls a distance H in 50 s when dropped on the surface of th...

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An object falls a distance H in 50 s when dropped on the surface of the earth. How long would it take for the same object to fall through the same distance on the surface of a planet whose mass and radius are twice that of the earth ? (Neglect air resistance.)a)35.4 gb)50.0 sc)70.7 sd)100.0 sCorrect answer is option 'C'. Can you explain this answer?
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