Kinematics, solving for theta

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1. Nov 1, 2014

Esoptron

1. The problem statement, all variables and given/known data
Use your calculated muzzle velocity (v_0) to determine the angle at which you would need to aim your launcher, fired from floor level, in order for your marble to land on a platform 10 cm above the floor, a horizontal distance of 50 cm from the launcher.

2. Relevant equations
x_f = x_0 + v_0 t + (1/2) a_x t^2
y _f= y_0 + v_0 t + (1/2) a_y t^2

3. The attempt at a solution
y _f= y_0 + v_0 t + (1/2) a_y t^2
10=Δt(sinΘ ⋅ v_0 - 1/2 ⋅ g ⋅Δt)
10=sinΘ ⋅ v_0 - 4.905(Δt)

x = x_0 + v_0 t + (1/2) a_x t^2
50=Δt(cosΘ ⋅ v_0 + 1/2 ⋅ 0 ⋅ Δt)
50=cosΘ ⋅ v_0 + 0 ⋅ Δt

Leads to this system of equations

v_0 ⋅ sinΘ - 4.905(Δt) - 10=0
v_0 ⋅ cosΘ + 0(Δt) - 50=0

- I don't know what the muzzle velocity is yet (This is prep for an upcoming lab) but I know that it is a constant and is the same for both equations so I'm leaving it as v_0.

- Looking at my final system of equations above, is it possible to get rid of Δt or express it in terms of constants, get theta by itself on the other side of the equal sign, and use that during the lab to see where the two equations would intersect to find an angle that would satisfy the question?

2. Nov 1, 2014

Simon Bridge

Yes. How would you normally solve simultaneous equations?

3. Nov 1, 2014

quantumtimeleap

Hi there!

Don't get scared of being a little creative and putting algebra to your advantage! :) Here's a hint: for the two equations put the trigonometric term ($\sin(\theta)$ or $\cos(\theta)$ ) on one side and the rest on the other side. Now divide the two equations. You should get $\tan(\theta) =$ something. Now that's an easy step away from the answer you're looking for! ^^

4. Nov 1, 2014

Simon Bridge

OK - so: $$\tan\theta = \frac{10+(4.905)\Delta t}{50-\Delta t}$$ ... now what?