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`        a particle is projected with an velocity an velocity u m/s making an angle​​​​​ “B” with horizontal. find average velocity of particle between the point of projection and the highest point of trajectory`
one year ago

Mohammed Kasib Siddiqui
119 Points
```							Let the highest point be P and point of projection be O. The the time of flight is given by T=2usinB÷g. Let the point Q be on the plane of projection such that it comes straight umder the pt. P .Join OPQ to form a triangle . Now PQ is the maximum height and OQ is half of range . Apply pythagoras theorem to find OP ie hypotenuse which will be our displacement.Now avg. velocity=displacement÷timeIe avg. velocity = OP÷ (usinB÷g)As it is half of the range hence the time of flight is T=usinB÷g
```
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