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 θ)^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

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