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Grade: 11
        
From the foot of a tower 90 m high a stone is thrown up so as to reach the top of the tower. Two seconds later another stone is dropped from the tower.The two stones meet at ?
 
9 months ago

Answers : (1)

Arun
22539 Points
							
Height of the building (d) = 90m
Let’s find the velocity of stone thrown up:
We know, v^2 – u^2= 2ad
Velocity of stone after reaching top of the building = 0
Acceleration is opposite to acceleration due to gravity = -9.8m/s^2
0- u^2= 2*(-9.8)*90
- u^2=-1764
Therefore, velocity of stone moving up= 42m/s
Distance at time t = h(t) =ut-1/2at^2
h(t) = 42t-4.9t^2 -----(1)
h1(T) = height of building – 1/2aT^2
T=t-2 (since second stone is thrown after two seconds)
H1= 90 – 4.9(t-2)^2
When the two stones meet, distance are same. So, Set h= h1
42t - 4.9t^2 = 90 -4.9t^2 +19.6t -19.6
42t-19.6t=90-19.6
22.4t = 70.4 => t = 3.14s 
Substitute the value of “t” in equation (1)
h =83.6m
So, the two stones meet at 83.6m 
 
 
 
9 months ago
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