Question

A stone is allowed to fall from the top of a tower 100m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25m/s. calculate when and where the two stones will meet.

Answer 1

Stone 1=u=0m/s
g=10m/s²
S=100-x
Time=T

Stone 2=u=25m/s
g=-10m/s²
s=x
Time=T

Equation =S=ut+1/2gt²

For stone 1=(100-x)=(0)t+1/2(10)t²
(100-x)=5t²..........1.)

For stone 2=x=(25)t+1/2(-10)t²
x=25t-5t²............2.)

x=25t-(100-x)..from 1.)
25t-100=0
25t=100
t=100/25
t=4s....(when???)

100-x=5(4)²
100-x=5(16)
100-x=80......3.)
x=100-80
x=20m(from ground)

25t-5t²
t(25-5t)=x
t(5t)=x
5t²=x
(100-x)=x... From 1.)
80=x...from 3.)
From top, x=80m....(where???)

Answer 2

_/\_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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