Question

Two objects are dropped from rest from the same height. Object A falls through a distance id="TexFormula1" title="d_A" alt="d_A" align="absmiddle" class="latex-formula"> during a time $t$, and object $B$ falls through a distance $d_B$ during a time $2t$. If air resistance is negligible, what is the relationship between $d_A$ and $d_B$?
Answer is $d_A=\frac{1}{4} *d_B$

Please show me how I can get the answer. Thank you.

Answer 1

Answer:

The answer to your question is given below

Explanation:

Since both object A and B were dropped from the same height and the air resistance is negligible, both object A and B will get to the ground at the same time.

From the question, we were told that object A falls through a distance to dA at time t and object B falls through a distance of dB at time 2t.

Remember, both objects must get to the ground at the same time..!

Let the time taken for both objects to get to the ground be t.

Time A = Time B = t

But B falls through time 2t

Therefore,

Time A = Time B = 2t

Height = 1/2gt^2

For A:

Time = 2t

dA = 1/2 x g x (2t)^2

dA = 1/2g x 4t^2

For B

Time = t

dB = 1/2 x g x t^2

Equating dA and dB

dA = dB

1/2g x 4t^2 = 1/2 x g x t^2

Cancel out 1/2, g and t^2

4 = 1

4dA = dB

Divide both side by 4

dA = 1/4 dB