Vikas TU
Last Activity: 8 Years ago
Let the speed be v and projection angle be thetha..
Then in x direcn. =>
30 = vcosthetha*t..............(1)
and
in y – direction,
40 = 0 + 0.5gt^2
t = 2root(2) seconds.
put in eqn. (1) we get,
vcosthetha = 30/2root(2) => 15/root(2)
v = 15/root(2)*cos(thetha)
For minimum v, denominator should be maximum and therfore, thetha should be minimum as it is a decreasing function.
Hence thetha = 0 degree and
V = 15/root(2) m/s
