# phy

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at what projection will the range of a projectile equal to 2 times its maximum height?

• phy -

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 -

thanks

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