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A particle is projected under gravity with velocity ✓(2ag) from a point at a height h above the level plane at angle x to it.Find the maximum range on the ground.

munazzah shaik , 7 Years ago
Grade 12
anser 1 Answers
Vikas TU

Last Activity: 7 Years ago

Dear Student,
The maximum range will depend on the horizontal velocity and the time it takes to reach ground.
v=√2ag
the displacement when it reaches ground is h
h=ut+at2/2
=>h=vsint-gt2/2
=>gt2-2vsint+2h=0
=>t=vsin+√(v^2  sin^2⁡-gh)/g
Range=vcos x t
=√2ag cos x (√2ag sin+√(〖2ag 〗^  sin^2⁡-gh))/g 
Here   = thetha.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

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