Question

An object is thrown from the height of 125 cm take g = 10 m/s. Find the ratio of distance covered by object in the 1 st and last 1 sec

• A. 1:9
• B. 4:9
• C. 4:4
• D. 2:9

Answer 1

The correct option is A 1:9

Given,

Initial velocity, $u=0$

Time period of free fall from height $125m$

$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\times 125}{10}}=5\sec$

Apply kinematic equation of motion

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

Ratio of displacement for $1^{st}$ sec and last $1\sec$

$\frac{s_1-s_0}{s_4-s_5}=\frac{\frac{1}{2}g(1^2+0)}{\frac{1}{2}g(5^2+4^2)}=\frac{1}{9}$

Hence , ratio is $1:9$