Question

a stone is allowed to fall from a tower of height 200 metre and at the same time another stone is projected vertically upward from the ground at a velocity of 20 metre per second calculate when and where the stone will meet​

Answer 1

Answer:

Explanation:

Stone thrown from the tower

u = 0

g = -10 m/s2

s = - (200- x)

s= ut +1/2 at2

-(200 - x) = 0 - 1/2 10 X t2 ................................................. (1)

For upward direction,

u = 20 m/s

g = -10 m/s2

s = +x

+x = 200 t - 1/2 10 t2 .............................................................(2)

Substracting 1 and 2, we get,

200 = 20t

t = 10s

x = 20 X 10 - 1/2 X 10 X 10 (square)

  = 200 - 500 m

  = 300 m

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Answer 2

The height at which the two stones will meet at a height (-80+20√160) m from the bottom of the tower and will meet after a time (-4 + √160) seconds.

Let the two stones meet at a height h from the from the bottom of the tower.

The distance covered by the stone dropped from the top of the tower is 200-h.

200-h = 1/2 gt²

=> 200-h = 5t² ------(1)

The distance covered by the stone thrown from the bottom of the tower is h.

h = 20 × t

=> h = 20t ------(2)

Putting the value of h from (1) in (2).

200 - 20t = 5t²

=> t² + 4t -40 = 0

=> t = (-4 + √160) seconds

h = 20t = 20 × (-4+√160) m