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`        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 ? `
one year ago

Arun
23742 Points
```							Height of the building (d) = 90mLet’s find the velocity of stone thrown up:We know, v^2 – u^2= 2adVelocity of stone after reaching top of the building = 0Acceleration is opposite to acceleration due to gravity = -9.8m/s^20- u^2= 2*(-9.8)*90- u^2=-1764Therefore, velocity of stone moving up= 42m/sDistance at time t = h(t) =ut-1/2at^2h(t) = 42t-4.9t^2 -----(1)h1(T) = height of building – 1/2aT^2T=t-2 (since second stone is thrown after two seconds)H1= 90 – 4.9(t-2)^2When the two stones meet, distance are same. So, Set h= h142t - 4.9t^2 = 90 -4.9t^2 +19.6t -19.642t-19.6t=90-19.622.4t = 70.4 => t = 3.14s Substitute the value of “t” in equation (1)h =83.6mSo, the two stones meet at 83.6m
```
one year ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions