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`        A stone dropped from the top of the tower travel 25m in the last second of its motion. If g is 10m/s, height of the tower is?`
2 years ago

Kruthik
119 Points
```							The answer is 45metresLet’s consider the last second of its journey.We have S=25m, a=g, t=1sThus, by using s=ut + 1/2at2, we get u=20ms-1 Now, during the time before the last second, let h be the distance covered. Here, S=h, a=g, v=20ms-1,u=0Therefore, by using V2=U2 +2aS, we have S=h=20m.Now, the total height is h+25 = 45meters. Hope that helped you. ✌️
```
2 years 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