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=vsint-gt2/2
=>gt2-2vsint+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)