A ball is launched from level ground with speed v0 at angle heta above horizontal. Neglect air resistance. What launch angle gives the maximum horizontal range, and what is that range?

Question

A ball is launched from level ground with speed v0 at angle 	heta above horizontal. Neglect air resistance. What launch angle gives the maximum horizontal range, and what is that range?

Accepted Answer

heta = 45°, range v0^2/g

Range on level ground is R = v0^2 \sin(2	heta)/g. Maximum occurs at 2	heta = 90°, so 	heta = 45°, giving R{\max} = v0^2/g.

Answer

heta = 60°, range v0^2 \sqrt{3}/g

Answer

heta = 45°, range v0^2/(2g)

Answer

heta = 30°, range v0^2/(2g)

Answer

heta = 45°, range v0^2/g

Vectors and Motion in Two Dimensions

Related questions