Question

A body of $1\text{kg}$ is thrown vertically upwards with an initial velocity $\left(u\right)=2\text{m/s}$, the height at which kinetic energy of the body is $(\frac{1}{4}{)}^{\text{th}}$ of its original value is (take $g=10{\text{m/s}}^2$)

• A. $0.15\text{m}$
• B. $0.25\text{m}$
• C. $0.3\text{m}$
• D. $0.35\text{m}$

Answer 1

The correct option is A $0.15\text{m}$

Given, initial velocity $\left(u\right)=2\text{m/s}$
Mass of body $\left(m\right)=1\text{kg}$
$\therefore$ Initial kinetic energy $\left(K_i\right)=\frac{1}{2}m{u}^2=\frac{1}{2}\times 1\times (2{)}^2=2\text{J}$
Final kinetic energy $\left(K_f\right)=\frac{1}{4}$ Initial kinetic energy $\left(K_i\right)$
From the law of conservation of energy,
$U_i+K_i=U_f+K_f$
$U_i=0\text{J};U_f=mgh\text{J}$
$\therefore K_i=U_f+K_f$
$U_f=K_i-K_f$
$=K_i-\frac{K_i}{4}=\frac{3K_i}{4}=\frac{3}{4}\times 2=\frac{3}{2}\text{J}$
$mgh=\frac{3}{2}\text{J}$
$\Rightarrow h=\frac{3}{2\times mg}=\frac{3}{2\times 1\times 10}=\frac{3}{20}=0.15\text{m}$