Question

A ball is released from height (h) &another from (2h). The ratio of the time taken by the two balls to reach the ground is what?

Answer 1

s = ut + 1/2at1 2
s=h
h = 4.9 t1 2
t1 = √h/4.9
t2 = √2h/4.9
t1:t2 = 1:√2

Answer 2

The time taken by the balls to reach ground is $\frac{1}{\sqrt{2}}$

Explanation:

Ball released from height h and height 2h simultaneously.

Those will have relation between them for the time taken to reach ground, which is to be calculated using the equation of motion.

The equation of motion used will be,

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

In both the cases, initial velocity u = 0 as balls released

$h_1=\frac{1}{2} g t_{1}^{2}$

$h_2=\frac{1}{2} g t_{2}^{2}$

On substituting the value of height, we get,

$h=\frac{1}{2} g t_{1}^{2} \text { and } 2 h=\frac{1}{2} g t_{2}^{2}$

$\frac{h_1}{h_2}=\frac{1}{2}=\frac{t_{1}^{2}}{t_{2}^{2}}$

$\frac{1}{2}=\frac{t_{1}^{2}}{t_{2}^{2}}$

$\frac{t_{1}}{t_{2}}=\frac{1}{\sqrt{2}}$