Question

Two identical objects A and B fall from rest from different heights to the ground. If object B takes twice as long as A to reach the ground, what is the ratio of the heights (A:B) from which the two objects fell? Neglect air resistance.

Answer 1

I believe this ratio is 4:1 due to the inverse square law

Answer 2

The ratio of heights from which the two different objects fell is $\boxed{1:4}$.

Explanation:

the two identical objects fall from rest. So, it means that the initial velocity of the object as it falls from the height is $0\text{ m/s}$.

The motion of the object as it falls from a height is the motion under the acceleration due to gravity of the earth. The mathematical expression for the distance covered by a object as it falls is:

$\boxed{s=v_{i}t+\dfrac{1}{2}gt^{2}}$                                              ...... (1)

Here, $s$ is the displacement of the object, $t$ is the time taken, $v_{i}$ is the initial velocity of object and $g$ is the acceleration due to gravity.

The object B takes twice as long as the time taken by the object A to reach the ground. Consider that the object A takes time $t'$ and the object B will take $2t'$ to reach the ground.

Substitute $0 \text{ m/s}$ for the initial velocity of the body.

The ratio of the height from which the objects fall can be obtained as:

$\dfrac{s_A}{s_B}=\dfrac{0+\dfrac{1}{2}g(t')^2}{0+\dfrac{1}{2}g(2t')^2}\\\dfrac{s_A}{s_B}&=\dfrac{1}{4}$

Therefore, the height from which the object B falls is four times the height from which the object A falls.

Thus, the ratio of heights from which the two different objects fell is $\boxed{1:4}$.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Motion in one dimension

Keywords:

two identical object, acceleration due to gravity, twice, time taken to fall, fall through height, ratio of heights, initial velocity, air resistance.