Two balls A and B are thrown simultaneously ,A vertically upwards with a speed of 20m/s from ground,and B vetically downwards from a height of 40 m with the same seed and along the same line of motion . At what points do the two balls collide ?

Suppose they meet at a height h from ground in time t. Their initial velocity of projection be u = 20 m/s

For the ball thrown upwards, (u and g are opposite)

h = ut - gt2 /2

For the ball thrown vertically downwards, (u and g are in the same direction)

40 - h = ut + gt2 /2

Add the two equations

40 = 2ut

t = 40/2u = 40/2x20 = 1 sec

h = 20 x 1 - 9.8 x 12 /2 = 20 - 4.9 = 15.1 m

They meet at height 15.1 m from the ground in 1 second.

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Because u is already given
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In the first case, as u and g are opposite, why isn't u negative and g taken as positive to use the same orientation of the x and y axis as the second case? Furthermore, from a common sense perspective, the first ball going upwards has to first go up, velocity becomes zero and then proceed downward to the original 40 m point which will take time. While the second ball is thrown downwards with an initial velocity of 20 m/s and is further accelerated by gravity. It is obvious they will reach only at the ground level and not in midair because the second ball while reach sooner. And the time it will take should be however long the first ball going upwards takes to reach the ground. 
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Please find this answer

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Wrong
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Very easy
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Simplest ans

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Hope this hepls

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15.1 m
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Suppose they meet at a height h from ground in time t. Their initial velocity of projection be u = 20 m/s

For the ball thrown upwards, (u and g are opposite)

h = ut - gt2?/2

For the ball thrown vertically downwards, (u and g are in the same direction)

40 - h = ut + gt2?/2

Add the two equations

40 = 2ut

t = 40/2u = 40/2x20 = 1 sec

h = 20 x 1 - 9.8 x 12?/2 = 20 - 4.9 = 15.1 m

They meet at height 15.1 m from the ground in 1 second.
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Here is the solution with explaination

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Ans is 15 m

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U1=20 m/s (upward)

U2=20(downward)

Let X be the distance from the ground where the collision takes place then

Distance from top of tower up to the collision point =40-X

If we suppose that collision took place after time T Then

X=20T-1/2×g×T^2……(1)(upward motion)

40-X=20T+1/2g×T^2….(2)(downward motion)

Adding (1) and (2) we get

40=40T

Which gives T=1 second

Putting value of T in (1) we get

X=20–1/2 ×9.8×1 ( taking :g= 9.8 m /s^2)

X=20–4.9=15.1 metre

So collision takes at 15.1 metres from ground or 40–15.1=24.9 metres from top.

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Is it helpful for you? If yes then ?? phod do.

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yes bro
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thanks for this easy question
 
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The balls will met at a height of 20m ...
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The food
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To mek sentes
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To go at home
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Gravitation
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15 is the answer
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X=20?1/2 ?9.8?1 ( taking :g= 9.8 m /s^2)

X=20?4.9=15.1 metre

So collision takes at 15.1 metres from ground or 40?15.1=24.9 metres from top.
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Takeg=9.8ms-2
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0 coll
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Answer

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