You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Physics

Law of Conservation of Energy

A stone of ma...

Question

A stone of mass 500 g is dropped from a certain height. When it is exactly at the midpoint of free fall, the kinetic energy possessed by it is 800 J. What is the height from where it is dropped ? (Take acceleration due to gravity of earth as $10 m s^{- 2}$ )

320 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

160 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

80 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

240 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $320 m$

When the stone is dropped to half of the height

$g a i n i n K . E = L o s s o f P . E 800 = m g h 800 = m g \frac{H}{2} H = Height from where the stone is dropped 800 = \frac{500}{1000} \times 10 \times \frac{H}{2} H = \frac{1600}{5} = 320 m H = 320 m$

Suggest Corrections

Similar questions

Q. We know that If a ball of mass 1 kg is kept at a height 10 m possess potential energy

m g h

i.e.

(1 \times 10 \times 10) = 100

J. (Taking Earth surface as reference, where

m

= mass of object,

g

= acceleration due to gravity,

h

= height of object from earth surface.)

what will be the kinetic energy of the ball falling freely from a height of 10 m when it reaches the ground? (take g = 10

m / s^{2}

and neglect air drag)

Q. The ball of mass 0.4 kg is dropped from some certain height .As it falls through 2 m. The kinetic energy acquired by it is.

Q. A stone is dropped on the moon from a height of

30 m

. The acceleration due to gravity on the moon is

\frac{g}{6}

. Determine the approximate time taken by the stone to fall on the surface of the moon.
(Take

g = 10 m / s^{2}

)

Q. Find the potenial energy possessed by a body of mass

4 k g

when it is at height of

10 m

above the ground. (Take

g = 10 m s^{- 2}

)

A body is dropped from a certain height at time $T = 0$ . It undergoes freefall and reaches the ground at time $T = t$ s. What is the expression for the height $h$ from which the body started its free fall?

Acceleration due to gravity is $g$ .

Join BYJU'S Learning Program

A stone of mass 500 g is dropped from a certain height. When it is exactly at the midpoint of free fall, the kinetic energy possessed by it is 800 J. What is the height from where it is dropped ? (Take acceleration due to gravity of earth as 10 ms−2)

The correct option is A 320 m When the stone is dropped to half of the height gain in K.E=Loss of P.E800=mgh800=mgH2H=Height from where the stone is dropped800=5001000×10×H2H=16005=320 mH=320 m

A stone of mass 500 g is dropped from a certain height. When it is exactly at the midpoint of free fall, the kinetic energy possessed by it is 800 J. What is the height from where it is dropped ? (Take acceleration due to gravity of earth as $10 m s^{- 2}$ )

The correct option is A $320 m$

When the stone is dropped to half of the height

$g a i n i n K . E = L o s s o f P . E 800 = m g h 800 = m g \frac{H}{2} H = Height from where the stone is dropped 800 = \frac{500}{1000} \times 10 \times \frac{H}{2} H = \frac{1600}{5} = 320 m H = 320 m$