Answer:
(i) The stones meet at 1.8 second
(ii) The point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.
Explanation:
(i)
First we consider the stone dropped by the girl. We have data:
Vi = Initial Velocity of Stone = 0 m/s (Since the stone was initially at rest)
t = Time Period
g = 10 m/s²
s₁ = Distance Covered by Stone
Using 2nd equation of motion, we get:
s₁ = Vi t + (0.5)gt²
s₁ = (0)(t) + (0.5)(10)t²
s₁ = 5t² ----- equation (1)
Now, we consider the stone throne vertically upward by the boy. We have data:
Vi = Initial Velocity of Stone = 25 m/s
t = Time Period
g = - 10 m/s² (negative sign due to upward motion)
s₂ = Distance Covered by Stone
Using 2nd equation of motion, we get:
s₂ = Vi t + (0.5)gt²
s₂ = (25)(t) + (0.5)(-10)t²
s₂ = 25t - 5t² ----- equation (2)
At, the point where both the stones meet, the sum of distances covered by both stones must be equal to the height of building (i.e 45 m).
s₁ + s₂ = 45
using values from equation (1) and equation (2)
5t² + 25t - 5t² = 45
25t = 45
t = 45/25
<u>t = 1.8 sec</u>
(ii)
using this value of of t in equation (2)
s₂ = (25)(1.8) - (5)(1.8)²
<u>s₂ = 28.8 m</u>
using this value of of t in equation (1)
s₁ = (5)(1.8)²
<u>s₁ = 16.2 m</u>
<u>Hence, the point at which the stones meet, is 28.8 m above the base of the building and 16.2 m below the top of the building.</u>
= 2.7 m/s