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
let the two stone meet at C after time t .
let AC = x
BC = 100-x
Taking vertical downward motion of stone .
u= 0 , a = 9.8m/s^2 ,S =x , t= t
As
S = ut + 1/2 at^2
x= 0 + 1/2 x 9.8 x t^2
x= 4.9 t^2 -------------(1)
Taking vertical upward motion of the stone thrown from B
u = 25 m/s
a= 9.8 m/s
S = ut + 1/2 at^2
100-x = 25t + 1/2 ( -9.8 )t^2
= 25t- 4.9 t^2 ----------------(2)
Add (1) and (2)
100 = 25 t
t= 4s
putting value in (1)
x= 4.9 x 16
= 78.4 m
in 4s and at a distance of 78.4 m below from top .
_/\_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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