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`        A projectile takes 4secs to pass through points A and B which are at a height 30.4 m above the ground. The Max height attained by the projectile is`
one year ago

Agent K
17 Points
```							Let h be max. Height, then Vy (y component of velocity) at h = 0Sy ( distance traveled vertically in those 2 seconds after reaching h max) = 0 + 1/2 g t^2 = 19.6mi.e 19.6 m above 30.4m => H max = 50m
```
one year ago
Agent K
17 Points
```							Left some spaces... Let h be max. Height, then Vy (y component of velocity) at h = 0Sy ( distance traveled vertically in those 2 seconds after reaching h max) = 0 + 1/2 g t^2 = 19.6mi.e 19.6 m above 30.4m => H max = 50m
```
one year ago
Geetha
10 Points
```							Max height H =30.4 +x Time taken to pass A and B = 4 secs= time of flightStamderd formula :4H=1/2gt^2                                  =4x=5*4^2                                       =x=20m  Therefore total height =30+20=50m(Take g=9.8 for more accuracy)
```
one year 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