At what angle (to the horizontal) must the projectile be fired to hit a target positioned at a range of 10.0 km away if the initial speed of the projectile v0 = 500 m/s?

I know I must find both the horizontal and vertical velocities, then use arctan, but have no idea how to do the horizontal velocity.

Question

At what angle (to the horizontal) must the projectile be fired to hit a target positioned at a range of 10.0 km away if the initial speed of the projectile v0 = 500 m/s?

I know I must find both the horizontal and vertical velocities, then use arctan, but have no idea how to do the horizontal velocity.

anonymous · Answer

Horizontal component of the velocity is 

Vx=500*Cos(alpha) 

you can find vertical component on your own. 

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anonymous · Answer

I'm still not sure of the steps to get the answer. The answer should be 11.8 degrees. The vertical velocity would be Vy = 500(sin)alpha. What do I do with that after? 

Someone said the equation should be this: 
Range = u^2Sin2θ/g

10*10^3 = 500^2*Sin2(θ)/9.8

= 11.54 degree

*Does anyone know how he got 10*10^3?

wolfe8 · Answer

10*10^3 is the range you want, that is 10 km written meters since the velocity given is in m/s.

anonymous · Answer

Oh duh! thank you

wolfe8 · Answer

You're welcome. It's helpful to convert everything into one unit you are comfortable working with before you start solving. It helps reduce mistake :)

anonymous · Answer

Thanks again, will try to remember that from now on. I try but keep forgetting. :)

anonymous · Answer

I am having trouble getting the right answer. It should be 11.54 degrees, I think I am using my calculator wrong or calculating wrong. 

After plugging everything in I get:
Theta = arcsin 1/2* (10*10^3 (20)/500^2)
Theta = arcsin .4 = 23.6 degrees. 

I am in degree mode, can't figure out what I am doing wrong?

wolfe8 · Answer

This is what I get when I solve for theta. $$\theta =\frac{ 1 }{ 2 } \sin^{-1} \frac{(10X10^{3})9.8 }{ 500^{2} }$$
Make sure you solve everything in brackets first, then multiply, divide, then arsin it. Then divide by 2

anonymous · Answer

Ah I think I was maybe maybe using the wrong formula. The one my professor gave me was Range = $$Range = V0^2\sin^2\alpha/2g$$

However I found on wikipedia the formula should be $$\alpha = 1/2 \sin^{-1} (gd/v0^2)$$, g being =10 (my prof allows us to use 10 for simplicity) and d = distance

So why 2g vs. g?

wolfe8 · Answer

Sorry I have to go now. Good luck.

anonymous · Answer

So with g = 10 it is 11.8 degrees. However I'm not sure I understand why his formula was different. Thanks though!