Supposly simple balistic motion prob

[assaftolko]

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

 

[voko]

This follows from sin x = sin (∏ - x).

 

[assaftolko]

This follows from sin x = sin (∏ - x).

wow you're right... so why do we say sin has a period of 2∏ if we can see that angles that are separated less than 2∏ have the same value?

 

[voko]

f(x) is said to be a-periodic if f(x) = f(x + a). For f(x) = sin x, this works when a = 2∏.

f(x) = f(a - x) is a different property, not periodicity.

 

[assaftolko]

f(x) is said to be a-periodic if f(x) = f(x + a). For f(x) = sin x, this works when a = 2∏.

f(x) = f(a - x) is a different property, not periodicity.

ok so what is this property? where does it come from?

 

[voko]

ok so what is this property? where does it come from?

I am not sure whether it has a special name. You could say that it comes from the definition the sine function on the unit circle. Then it is obvious that the ordinate at some angle a equals the ordinate at (∏ - a).

Another way to see this is by using the identity sin (∏ - a) = sin ∏ cos a - cos ∏ sin a = sin a, because sin ∏ = 0, and cos ∏ = -1.

 

[assaftolko]

I am not sure whether it has a special name. You could say that it comes from the definition the sine function on the unit circle. Then it is obvious that the ordinate at some angle a equals the ordinate at (∏ - a).

Another way to see this is by using the identity sin (∏ - a) = sin ∏ cos a - cos ∏ sin a = sin a, because sin ∏ = 0, and cos ∏ = -1.

tnx a lot!