 # 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.

(i) When the stone from the top of the tower is thrown,

Initial velocity u = 0

Distance travelled = x

Time taken = t

Therefore,

Height, h = (100 – x)

Initial velocity, u = 0
Time, t =?

g = 10 m/s2 (As the stone is falling down)
We know that,

h= ut+1/2gt2

Substituting the values in above equation
(100 – x) = 0 × t +1/2× 10 × t2

(100 – x) = 5t2 (i)
(ii)Now, for stone projected vertically upwards:

Height, h = x
Initial velocity, u = 25 m/s

Time, t =?
g = -10 m/s2 (Because the stone is going up)

We know that,

s= ut+1/2gt2

(100 – S) = 25t + 1/2 (-10) t²

= 25t – 5t²

On adding the above equations We get

100 = 25t

or t = 4 s

After 4sec, two stones will meet

From (a)

x = 5t2 = 5 x 4 x 4 = 80m.

Putting the value of x in (100-x)

= (100-80) = 20m.

This indicates that after 4seconds, 2 stones meet a distance of 20 m from the ground.

(i) When the stone from the top of the tower is thrown,

Initial velocity u = 0

Distance travelled = x

Time taken = t

Therefore,

Height, h = (100 – x)

Initial velocity, u = 0
Time, t =?

g = 10 m/s2 (As the stone is falling down)
We know that,

h= ut+1/2gt2

Substituting the values in above equation
(100 – x) = 0 × t +1/2× 10 × t2

(100 – x) = 5t2 (i)
(ii)Now, for stone projected vertically upwards:

Height, h = x
Initial velocity, u = 25 m/s

Time, t =?
g = -10 m/s2 (Because the stone is going up)

We know that,

s= ut+1/2gt2

(100 – S) = 25t + 1/2 (-10) t²

= 25t – 5t²

On adding the above equations We get

100 = 25t

or t = 4 s

After 4sec, two stones will meet

From (a)

x = 5t2 = 5 x 4 x 4 = 80m.

Putting the value of x in (100-x)

= (100-80) = 20m.

This indicates that after 4seconds, 2 stones meet a distance of 20 m from the ground. (140) (81)