a stone falls from the top of a building 200m high and at the same time another is projected vertically upwards with a velocity of 50m/sec ,then the two will meet
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A stone is dropped from the top of a tower that is 200 meters tall. At the same time, another stone is projected upward from the ground with the velocity of 50 meters/second. How do I find where and when both the stones meet?
8 Answers
Abdullah Safarini
Abdullah Safarini, In love with mathematics and philosophy.
Answered Jul 16, 2019
Write two equation for the displacements of both. In terms of time and other factors.
When they meet. Their displacements from the ground are equal. Also, time spent traveling is equal.
So 200 minus the displacement of the stone dropped equals the displacement of the other stone. Now substitute the displacement in the dropped stone equation with (200 minus the displacement of the other stone)
You have two equations now, two unknowns, which are time and displacement. But time and displacement for both equations are equal now, since we write the displacement of the first stone in terms of the di...
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Suresh Verma
Suresh Verma, M.Sc.(Physics) Physics, Patna Science College, Patna (1965)
Answered Apr 5, 2018 · Author has 2k answers and 1.6m answer views
to get to the answer /real answer
one must find out the time taken by the body sent upward to meet the top body (as if the body was not dropped)
so if you divide the height 200m by the velocity 50m/s then one gets 4 sec.
if you do all equations as done by another responder for accelerated motion ultimately you will get the same answer.
you can ask why you are getting the correct time interval even if g was put off.
it so happens that the acceleration produced each sec to the top body (dropped)is also a deceleration produced to the velocity of bottom particle, so net effect of g gets adjusted.
now...
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Muthusamy Piramanayagam ( முத்துசாமி பிரமநாயகம்), M.Sc. Retired as Head of Department of Physics. Government of Tamilnadu, lndia.
Updated Aug 29, 2018 · Author has 4.3k answers and 1.9m answer views
Since both are subjected to gravity equally, one ignores gravity, And the stone does not fall and is at rest at a height of 200 m.
The second meets that stone in 200/50 = 4 second.
Coming to real situation , the dropped one falls a distance of 1/2 g t^2.
4.9 * 4 *4 = 78.4 m from the top or
200- 78.4 from the bottom. They meet at this position .
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Abhimanyu Kumar (अभिमन्यु कुमार)
Abhimanyu Kumar (अभिमन्यु कुमार), Ilahia college of engineering kerala, India
Answered Jul 14, 2018 · Author has 69 answers and 180.9k answer views
Since,S=ut+1/2 at^2.
In case of free fall S=ut+1/2gt^2 eqn…1
Two cases arises here,
Case:1- stone is dropped from the tower,
given
S1=200 meters , u=0
So that, S1=ut+1/2 at^2 since it's case of free fall
So, S=ut+1/2gt^2 => S=1/2gt^2 eqn..2
Case:2 stone projected upward
S2=50t-1/2 gt^2. eqn ..3
Total distance S=S1+S2
Or, 200=S1+S2
Or, 200=1/2 gt^2+50t-1/2gt^2
=> 200=50t=> t=4sec.
Put t=4 in eqn 2.
We have, S1=0+1/2gt^2 => 1/2×9.81×4^2= 78.48m.
And S2=50×4–1/2×9.81×4×4
=200–78.48=121.52m
Both stones meet from ground in air at the distance of 121.52m in 4s.
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