A body is dropped from a height H. if gravity ceases to act after the second, the time of fall of the body is?

Open in App

Solution

if the gravity ceases after a second

there will be no more acceleration due to gravity but within that one second it had got some

velocity downwards and continues to move downwards with that velocity (neglecting air resistance)

so

velocity after one second

v = u + at , u = 0 , t = 1 sec , a= 9.81

v = 9.811 = 9.81 m / sec

now distance traveled in that one second

s = ut + 1/2 a t^2

s= 1/29.81*1^2

s= 4.9m

now the distance to be traveled is H - 4.9

we know velocity = displacement / time

time = dispalcement / velocity

time = (H - 4.9)/ 9.81

so the total time for the fall of the body is = 1 + (H- 4.9)/9.81

by taking LCM

= (9.81 + H - 4.9) / 9.81

= (4.9 +H) / 9.81

Suggest Corrections

Similar questions

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}

)

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/ $s^{2}$ )

A body falls for $5 s$ from rest. If the acceleration due to the gravity of earth ceases to act, The distance travels in the next $3 s$ is.

Q. A body is dropped from a height

h

number acceleration due to gravity

g

. If

t_{1}

and

t_{2}

are time intervals for its fall for first half and the second half distance, the relation between them is :

Q. A body falls from a height of

200 m

. If gravitational attraction ceases after

2

s, further time taken by it to reach the ground is

(g = 10 m s^{- 2})

Join BYJU'S Learning Program