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 (acceleration due to gravity = 9.8 ms-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 (acceleration due to gravity = 9.8 ms-2) (A) 5 m (B) 2.5 m (C) 10 m (D) 12.5 m

Tardigrade Admin · Accepted Answer

Given, m=5 kg and KE =490 J By the law of conservation of energy (1/2) m u2 =(1/2) m v2+m g h 490 =245+5 × 9.8 × h h =(490-245/5 × 9.8) h =(245/49) =5 m

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 (acceleration due to gravity = $9.8 m s^{- 2}$ )

5 m

2.5 m

10 m

12.5 m

Given, $m = 5 k g$
and $K E = 490 J$
By the law of conservation of energy
$\frac{1}{2} m u^{2} = \frac{1}{2} m v^{2} + m g h$
$490 = 245 + 5 \times 9.8 \times h$
$h = \frac{490 - 245}{5 \times 9.8}$
$h = \frac{245}{49}$
$= 5 m$

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 (acceleration due to gravity = 9.8ms−2)

5 m

2.5 m

10 m

12.5 m

Given, m=5kg and KE=490J By the law of conservation of energy 21​mu2=21​mv2+mgh 490=245+5×9.8×h h=5×9.8490−245​ h=49245​ =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 (acceleration due to gravity = $9.8 m s^{- 2}$ )

Given, $m = 5 k g$
and $K E = 490 J$
By the law of conservation of energy
$\frac{1}{2} m u^{2} = \frac{1}{2} m v^{2} + m g h$
$490 = 245 + 5 \times 9.8 \times h$
$h = \frac{490 - 245}{5 \times 9.8}$
$h = \frac{245}{49}$
$= 5 m$