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Can anybody please answer the question in the attached image

Can anybody please answer the question in the attached image 

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1 Answers

freaky1 town
35 Points
5 years ago
Both the inclines are inclined at 30° .
The projected particle makes 90° with incline ,resolving this angle with respect to ground we get that it makes 30° with verticle. Now resolving the velocity component we get verticle velocity = ucos(30)= u√3/2 
For max height we know verticle velocity must go 0 therefore time required to reach max height=u√3/2g (v=u +at)
We know time of ascend = time of decent
Therefore time of flight = 2* time to reach max height.
Time of flight =u√3/2g *2= u√3/g..
 

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