Posted by ami on Sunday, September 5, 2010 at 12:35pm.
at what projection will the range of a projectile equal to 2 times its maximum height?

phy  bobpursley, Sunday, September 5, 2010 at 12:50pm
The time of flight is given by 2Vi sinTheta/g
Range is then Vi*t*costheta=2Vi^2sinThetaCosTheta/g
The maximum height is given by H = (v sin Î¸)^2 / (2 g).
work all those out yourself to verify true.
Now, Range=2Height
2Vi^2sinThetaCosTheta/g =2(Vi*sinTheta)^2/2g
reducing that,
sinTheta/cosTheta=2
theta=arctan 2
check my math. 
phy  ami, Sunday, September 5, 2010 at 1:11pm
thanks