Question

A stone is thrown vertically upward with a velocity 20 m/s. At same time another stone is dropped from the top of a tower of height 50 m
in the same line motion of travel .When and where they will meet?​

Answer 1

Answer:

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A stone is thrown vertically upward with a velocity of 20m/s. At the same time and 20m vertically above, a second stone is allowed to fall. After what time and height do they collide?

Let the ball at the ground be A and that at the top be B.

Assume that you see the motion of A while sitting at B.

You(B) are obviously at rest w.r.t. yourselves. However you(B) and A have the same acceleration in the same direction. Thus acceleration of A w.r.t. you(B) is 0.

Thus, to you, it will appear as if A is travelling towards you with an uniform speed of 20m/s.

Dividing the distance 20m with that speed, we get that A reaches you(B) in 1s.

Now, you have considered these motions and time by assuming you were stationed at B. However, the time taken for A to meet B shouldn't be dependent on which reference frame you assume. Thus, time taken for A and B to collide is 1s.

Since B falls freely, he covers a distance of (assuming g = 10m/s²)

Thus, they meet at a height of (20–5)m = 15m from the ground.

Answer 2

When stone 1 is thrown up with an initial velocity of 19.6m/s

, the final velocity of the ball after2 seconds is given by the motion equation,

v=u+at

,where v is the final velocity of the stone after 2 seconds.

⇒v=19.6+(−9.8)×2

(Since when the stone is thrown up, acceleration is due to the downward gravitational force)

⇒v=0m/s

So , the maximum height that stone one can reach can be identified using second equation of motion,

s=ut+12gt2

, where s is the maximum height the stone can reach.

⇒s=(19.6×2)+12(−9.8×4)

⇒s=19.6m

Thus stone 1 can reach a maximum height of 19.6 meters in 2 seconds from ground when thrown at19.6m/s

. In order for the second stone to reach the first stone at 19.6 meters, when it is thrown at 9.8m/s

, it will take another 2 seconds to reach the point of maximum height.

If a stone to reach a point against gravity takes a time period, then the time period for it to come back to its normal position will also be the same. Hence, stone 1 will take 2 seconds to drop ground from 19.6 meters. As stone 1 starts to move towards ground, stone 2 starts its acceleration . Thus when stone 2 is at 19.6 meters, stone 1 will land on the ground and hence stone 2 after a period of 4 seconds. Thus both the stones will meet on the ground after a period of 4 seconds.