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.
check my math.
phy - ami, Sunday, September 5, 2010 at 1:11pm