Question

Attention!



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

Answer 1

Answer:-

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⏩S=>distance travelled by stone dropped from tower

⏩T=>time when both meet

⏩(100-S)=>distance travelled by projected stone

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For stone dropped from tower

-S=>0+1/2(-10)^2

S=>5t^2. (eq1)

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for stone projected upward

(100-S)=>25t+1/2(-10)t^2

=>25-t^2. (eq2)

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Adding eq1 and eq2

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t=>4s

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put the value in eq1,we get S=>80m

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thus the time when they meet 4s and both will meet at a distance of 80m

_________hope it helps

Answer 2

SOLUTION:-

Given:

⚫Height of the tower=100m.

⚫g= 10m/s²

⚫Velocity of projection of ground projectile= 25m/s

⚫Velocity of projection of the tower projectile= 0

Let the particles meet at time (t) and at height (h):

Thus,

Height covered by the projectile projected from ground:

$h = 25t - \frac{1}{2} g {t}^{2} \\ \\ = > h = 25t - \frac{1}{2} \times 10 \times {t}^{2} \\ \\ = > h = 25t - 5 {t}^{2} .............(1)$

Distance covered by the projectile thrown down in time (t) will be:

Since it falls from rest u= 0

$= > 100 - h = \frac{1}{2} g {t}^{2} \\ \\ = > 100 - (25t - 5 {t}^{2} ) = 5 {t}^{2} (by \: 1) \\ \\ = > 25t = 100 \\ \\ = > t = \frac{100}{25} \\ \\ = > t = 4sec.$

Thus the particles meet each other at 4seconds.

Height at which the particles meet is h,

So,

Given t= 4sec.

=) h= 25t - 5t²

=) h= 25(4)- 5(4)²

=) h= 100 - 80

=) h= 20m

Hence,

Two stones meet is 20m.

Hope it helps ☺️❤️