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# explain

Let us consider the first stone is dropped when balloon is at a position A . Let vA be its velocity.

After 2 seconds the balloon reaches the position B and its speed at B is vB

vB = vA + ( a × t )= vA + (0.2 × 2 ) = vA + 0.4    ......................................(1)

where a = 0.2 m/s2 is acceleration of balloon

In 2 seconds , balloon has travelled a distance h  that is determined from following equation

vB2 = vA2 + ( 2 a h )

h = ( vB2 - vA2 ) / ( 2 a ) = [ ( vA+0.4 )2 - vA2 ] / ( 2 × 0.2 )

h = ( 0.8 vA + 0.16 ) / 0.4  = 2 vA + 0.4  ...................................(2)

At position A , when a stone is dropped means , stone moves upwards with speed vA

Similarly, At position B , when a stone is dropped means , stone moves upwards with speed vB

If it is required to find the distances between two stones 1.5 s after dropping second stone,
then stone dropped at A has travelled 3.5 s and stone dropped at B has travelled only 1.5 s.

Let us find the distance by taking positio-A as reference point

If hA is the distance travelled by first stone, then

hA = ( vA t ) - [ (1/2) g t2 ] = ( 3.5 vA ) - [ 4.9 × 3.5 × 3.5 ]

hA = ( 3.5 vA ) - 60.025

If hB is the distance travelled by second stone, then

hB = h + ( vB t ) - [ (1/2) g t2 ]

Let us substitute h from eqn.(2)  and vB from eqn.(1)

hB  = ( 2 vA + 0.4  ) + [ 1.5 ( vA +0.4) ] - [ 4.9 × 1.5 × 1.5 ]

hB = ( 3.5 vA ) - 10.025

Distance between two stones is calculated as

hB - hA = ( 3.5 vA ) - 10.025 - [ ( 3.5 vA ) - 60.025 ]

hB - hA = 50 m

Answered by Expert 24th June 2022, 11:34 PM
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Tags: distance