A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Where are when will the two stones meet? Chapter - Motion & Kinematic Formulas Class IX
Answers
ince,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.
_/\_Hello mate__here is your answer--
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⚫Let the two stones meet after a time t.
CASE 1 :-When the stone dropped from the tower
u = 0 m/s
g = 9.8 ms−2
Let the displacement of the stone in time t from the top of the tower be s.
From the equation of motion,
s = ut + 1/2gt^2
⇒s = 0 × + 1/2× 9.8 ×t ^2
⇒ s = 4.9t^2 …………………… . (1)
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CASE 2 :--When the stone thrown upwards
u = 25 ms−1
g = −9.8 ms−2(upward direction)
Let the displacement of the stone from the ground in time t be '
Equation of motion,
s' = ut+ 1/2gt^2
⇒s′ = 25 × − 1/2× 9.8 × t^2
⇒s′ = 25 − 4.9t^2 …………………… . (2)
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Given that the total displacement is 100 m.
s′ + s = 100
⇒ 25 − 4.9t^2 + 4.9t^2 = 100
⇒ t =100 /25 = 4 s
The falling stone has covered a distance given by (1) as = 4.9 × 4^2 = 78.4 m
Therefore, the stones will meet after 4 s at a height (100 – 78.4) = 20.6 m from the ground.
I hope, this will help you.☺
Thank you______❤
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