(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 g = 10 m/s2 (As the stone is falling down) h= ut+1/2gt2 Substituting the values in above equation (100 – x) = 5t2 (i) Height, h = x Time, t =? 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.
Time, t =?
We know that,
(100 – x) = 0 × t +1/2× 10 × t2
(ii)Now, for stone projected vertically upwards:
Initial velocity, u = 25 m/s
g = -10 m/s2 (Because the stone is going up)
(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.
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