A body is dropped from a height 39.2 m. After it crosses half distance, the acceleration due to gravity ceases to act. The body will hit the ground with velocity (Take $g = 10 m s^{- 2} .$ ):

19.6 m s^{- 1}

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

20 m s^{- 1}

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

1.96 m s^{- 1}

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

196 m s^{- 1}

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

Open in App

Solution

The correct option is A $19.6 m s^{- 1}$
Suppose v be the velocity of the body after falling through half the distance. Then
$S = \frac{39.2}{2} = 19.6 m,$ $u = 0$ and $g = 9.8 m s^{- 2}$
$v^{2} = u^{2} + 2 g h = 0^{2} + 2 \times 9.8 \times 19.6$ $v = 19.6 m s^{- 1} .$
When the acceleration due to gravity ceases to act, the body travels with the uniform velocity of $19.6 m s^{- 1}$ .

Suggest Corrections

Similar questions

A body is dropped from a height of 39.2 m. After it crosses half the distance, the acceleration due to gravity ceases to act. The body will hit the ground with velocity (Take g=9.8 m/ $s^{2}$ )

Q. A body is dropped from a height

39.2 m

. After it crosses the half distance, the acceleration due to gravity ceases to act. Then the body will hit the ground with a velocity of (Take

g = 9.8 m s^{- 2}

)

Q. A body is dropped from a height of 39.2 m. After it crosses half distance, the acceleration due to gravity ceases to act. The body will hit the ground with K.E. (if m = 2kg) :-

Q. A body is projected from the ground at an angle of

30^{o}

with the horizontal with a velocity of

20 {m s}^{- 1}

. The magnitude of the average velocity of the body during its ascent is: (Acceleration due to gravity

= 10 {m s}^{- 2}

)

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

10 {m s}^{- 1}

and another body is projected simultaneously from the same point with a velocity of

20 {m s}^{- 1}

at an angle of

\frac{π}{6}

with the horizontal. The distance between the two bodies after one second from the time of projection is (Acceleration due to gravity is

10 {m s}^{- 2}

)

Join BYJU'S Learning Program

A body is dropped from a height 39.2 m. After it crosses half distance, the acceleration due to gravity ceases to act. The body will hit the ground with velocity (Take g=10ms−2.):

A body is dropped from a height 39.2 m. After it crosses half distance, the acceleration due to gravity ceases to act. The body will hit the ground with velocity (Take $g = 10 m s^{- 2} .$ ):