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A parachutist drops first freely from an airplane for 10s and then his parachute opens out.Now he descends with a net retardation of 2.5 ms-2.If he bails out of the plane at a height of 2495m and g=10ms-2, his velocity on reaching the ground will be
A parachutist drops first freely from an airplane for 10s and then his parachute opens out.Now he descends with a net retardation of 2.5 ms-2.If he bails out of the plane at a height of 2495m and g=10ms-2, his velocity on reaching the ground will be

```
3 years ago

Arun
25768 Points
```							 The initial velocity of the parachutist is zero now V=u+gt V=0+(-10)10 V=-100 (-sign as the velocity is in downward direction) Now S=ut+1/2ut^2 so S=0+1/2(-10)*100 S=500 Now H=2495-500 =1995 NOw v^2=u^2+2ah v^2=10000-2*2.5*1995 v^2=25 v=5 so his velocity on reaching the ground is 5m/s
```
3 years 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