- #1
assaftolko
- 171
- 0
At what angle relative to the horizon should you throw a ball at V0=15 m/s (from the ground of course) in order for it to hit back the ground a distance of 21m away?
I found that 2θ=66.16+2∏k, and so θ=33.08+∏k, but the truth is that θ=56.92 is also a correct answer! How could I get this answer from the equations??
x(t)=Vocosθt
y(t)=0 at hit so: 0=Vosinθt-0.5gt^2 ... t(impact) = 2V0sinθ/g or 0
back to x: 21=(2sinθcosθVo^2)/g = sin2θ*V0^2 / g and so sin2θ=21g/V0^2
I found that 2θ=66.16+2∏k, and so θ=33.08+∏k, but the truth is that θ=56.92 is also a correct answer! How could I get this answer from the equations??
x(t)=Vocosθt
y(t)=0 at hit so: 0=Vosinθt-0.5gt^2 ... t(impact) = 2V0sinθ/g or 0
back to x: 21=(2sinθcosθVo^2)/g = sin2θ*V0^2 / g and so sin2θ=21g/V0^2