Question

A ball takes t second to fall from a height h₁ and 4t seconds to fall from h₂. then$\frac{h_1}{h_2}$ is

• A. 2
• B. 4
• C. 0.5
• D. 0.0625

Answer 1

The correct option is D

0.0625

Step 1: Given:

1. A ball takes t second to fall from a height h₁.
2. A ball takes 4t seconds to fall from h₂.

Step 2: By using newton's second equation of motion:

$s=ut+\frac{1}{2}a{t}^2$

Where,

s is distance traveled, u is initial velocity, a is a constant acceleration, and t is time.

Now, we know that the initial velocity will be zero as the ball falls from rest and acceleration will be equal to the acceleration due to gravity(g).

Step 3: Finding heights h₁and h₂

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ s=h_1,u=0,a=g,t=t_1 \end{gathered}$$

${\mathrm{h}}_1=\frac{1}{2}{{\mathrm{gt}}_1}^2\Rightarrow h_1\propto {t_1}^2$

Similarly for height h₂,

${\mathrm{h}}_2=\frac{1}{2}{{\mathrm{gt}}_2}^2\Rightarrow h_2\propto {t_2}^2$

Finding the ratio of $\frac{h_1}{h_2}$:

$\frac{h_1}{h_2}={\left(\frac{t_1}{t_2}\right)}^2={\left(\frac{t}{4t}\right)}^2=\frac{1}{16}=0.0623\approx 0.0625.$

Hence, the correct option is (D).