Arun
Last Activity: 6 Years ago
a)initial energy = 1/2 m v^2
work done by air drag is f h
potential energy at maximum height h is mgh ( or the work done by/against gravirty)
1/2mv^2 = fh + mgh
h = mv^2 /2(f+mg) or writing interms of W = mg , replace m on top with W/g and mg in the debominator with W you will get the result
b) initial kinetic energy = 1/2mv^2
final energy = 1/2mV2^2 where v2 is the final velocity
work done by f is 2fh
1/2mv^2 = 1/2mV2^2 + 2fh since there is no change in potential energy
1/2 mv2^2 = 1/2mv^2 -2fh
W v2^2 = Wv^2 -4fgh
v2^2 = v^2 - 4gf/W * h
but we know h = v^2/(2g(1+(f/W))) = Wv^2/(2g(f+W)
v2^2 = v^2 - 4gf/W * Wv^2/(2g(f+W))
v2^2 = v^2 - 2f * v^2/((f+W)) = v^2 [(f+W -2f)/(W+f)] = v^2 (W-f)/(W+f)
v2 = v^2 √((W-f)/(W+f))
