A body of mass 5 kg is thrown vertically up with a kinetic energy of 490 J . The height at which the kinetic energy of the body becomes half of the original value is: (take g as 9.8 m s- 2 )

Question

A body of mass 5 kg is thrown vertically up with a kinetic energy of 490 J . The height at which the kinetic energy of the body becomes half of the original value is: (take g as 9.8 m s- 2 ) (A) 12.5m (B) 10m (C) 2.5m (D) 5m

Tardigrade Admin · Accepted Answer

According to the law of conservation of energy (1/2)mu2=(1/2)mv2+mgh 490=245+5× 9.8× h h=(245/49)=5m

Q. A body of mass $5 k g$ is thrown vertically up with a kinetic energy of $490 J$ . The height at which the kinetic energy of the body becomes half of the original value is : (take $g$ as $9.8 m s^{- 2}$ )

$12.5 m$

$10 m$

$2.5 m$

$5 m$

According to the law of conservation of energy
$\frac{1}{2} m u^{2} = \frac{1}{2} m v^{2} + m g h$
$490 = 245 + 5 \times 9.8 \times h$
$h = \frac{245}{49} = 5 m$

Q. A body of mass 5kg is thrown vertically up with a kinetic energy of 490J . The height at which the kinetic energy of the body becomes half of the original value is : (take g as 9.8ms−2 )

12.5m

10m

2.5m

5m