A body of mass $2$ kgf is projected vertically upwards with a velocity of $2$ m/s. The K.E. of the body just before striking the ground is:-

Open in App

Solution

At the same horizontal level, the velocity of the body is same.
So, the body strikes the ground with speed $= 2 m / s$
Kinetic energy $K . E = \frac{1}{2} m v^{2} = \frac{1}{2} \times 2 \times 2^{2} = 4 J$

Suggest Corrections

Similar questions

Q. a body of mass 2 kg is projected upwards with a velocity 4 metre per second. The kinetic energy of the body just before striking the ground is what?
answer:16J
please explain the answer in detail.

A body of mass $5 k g$ is thrown vertically upwards from ground with a velocity of $20 m s^{- 1}$ . What will be its momentum just before hitting the ground? Assume upward direction as positive.

Q. A body is projected vertically upwards with a velocity of

19.6 m / s

. The total time for which the body will remain in the air is (Take

g = 9.8 m / s^{2}

)

Q. A body is projected vertically upwards with a velocity of 10 m/s. The maximum height reached by the body is:

A body of mass $20 k g$ is dropped from a height of $2 m$ . If $g$ is taken to be equal to $10 m s^{- 2}$ , the kinetic energy of the body just before striking the ground will be:

Join BYJU'S Learning Program

A body of mass 2kgf is projected vertically upwards with a velocity of 2m/s. The K.E. of the body just before striking the ground is:-

At the same horizontal level, the velocity of the body is same. So, the body strikes the ground with speed =2 m/sKinetic energy K.E=12mv2=12×2×22=4 J

A body of mass $2$ kgf is projected vertically upwards with a velocity of $2$ m/s. The K.E. of the body just before striking the ground is:-

At the same horizontal level, the velocity of the body is same.
So, the body strikes the ground with speed $= 2 m / s$
Kinetic energy $K . E = \frac{1}{2} m v^{2} = \frac{1}{2} \times 2 \times 2^{2} = 4 J$