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 grouhnd with a velocity of 25m/s. Calculate when and where the 2 stones will meet. (take g= 10m/s squere)

Answer 1

let the stones meet after 't' sec at' x' m below the top of a tower.
In case 1,
u=0
h=x
g=10m/s^2
h=ut+1/2gt^2
h=5t^2___*
In case 2,
u=25m/s
h=100-x
g=-10m/s^2 (upward)
h=ut+1/2gt^2
(100-x )=25t-5t^2_____#
substituting the value of *in #,
100-5t^2=25t-5t^2
100=25t
t=4sec
now substituting the value of the in*
x= 80m
thanks....... hope so u will understand this

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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