Question

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. Calculate when and where the two stones will meet?​

Answer 1

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$\huge\mathfrak{Answer:-}$

⏩ Suppose the two stones meet at a height x from the ground, after time t from the start.

* For the downward motion of stone A:-

u= 0, g= -10 ms^-2, s= (-100-x)

As , s= ut + 1/2 gt²

Therefore,

-(100-x) = 0 - 1/2 × 10 × t² .... → (1)

* For the upward motion of stone B:-

u= +25 m/s, g= -10 m/s², s= +X

As, s= ut + 1/2gt²

Therefore,

+ x= 25 t -1/2 × 10 t² ....→ (2)

Now , on subtracting (1) from (2), we get,

100 = 25 t

or

t= 4s

From (2), x= 25 × 4 -1/2 × 10 × (4)²

= 100 - 80

= 20 m.

→ Hence, the two stones meet after 4s at a height of 20 m from the ground or 80 m from the top.

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Answer 2

Answer:

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