# a stone is allowed to fall from the top of a tower of height 200m and at the same time an another stone is projected vertically upwards from the ground with a velocity of 50 metre per second calculate where and when the two stones will meet. ( take g = 10 metre per second square)

### Asked by jy3037485 | 21st Mar, 2021, 11:05: AM

Expert Answer:

### The time taken by the body sent upward to meet the top body

is 200/50 = 4 seconds.

The acceleration produced each second to the top body which is dropped is also a deceleration produced to the velocity of stone thrown upward.

4.9 x 4 x 4 = 78.4 m

From the top 200 - 78.4 = 121.6 m

Thus,
Both stones meet at 121.6 m

is 200/50 = 4 seconds.

The acceleration produced each second to the top body which is dropped is also a deceleration produced to the velocity of stone thrown upward.

4.9 x 4 x 4 = 78.4 m

From the top 200 - 78.4 = 121.6 m

Thus,

### Answered by Shiwani Sawant | 23rd Mar, 2021, 11:37: PM

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