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`        a body is thrown vertically upwards and takes 5 seconds to reach maximum height . the distance travelled by the body will be same in`
6 months ago

raosuresh623
13 Points
```							V=u+gt, here at top v=0,  g is downward and a is upward ie. a=-g. 0= u-gt u=gt u=10×5u=50m/sFor distance V^2-u^2=-2gh0-2500=-2×10×ss=125mThat is answer
```
6 months ago
Arun
23330 Points
```							Dear student 1st and 10th second is correct answer.using g = 10 m/sec^2 : Vo = g*t = 10*5 = 50 m/sec from t = 0 to t = 1 sec h = ((Vo*1-5*1^2)-0) = (50-5) = 45 m (up) from t = 9 to t = 10 sec h = ((Vo*10-5*10^2)-(Vo*9-5*9^2) h= (0-450+405) = -45 m (down)
```
6 months ago
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### Course Features

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