What Angle Should You Aim the Launcher to Hit a Target 10 cm Above the Floor?

[Esoptron]

Homework Statement

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.

Homework 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

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?

 

[Simon Bridge]

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?

Yes. How would you normally solve simultaneous equations?

 

[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! ^^

 

[Simon Bridge]

Here's a hint: for the two equations put the trigonometric term (sin(θ) or cos(θ) ) on one side and the rest on the other side. Now divide the two equations. You should get tan(θ)= something.

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

 

[HalfLostAndSearching]

Yes, it is possible to eliminate Δt and solve for Θ in terms of constants. First, we can rearrange the equations to get Δt by itself:

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

Since Δt is the same in both equations, we can set the two expressions equal to each other:

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

This is undefined, but we can take the limit as Δt approaches 0 to find the value of Θ:

lim Δt→0 (v_0 ⋅ sinΘ - 10)/4.905 = lim Δt→0 (50 - v_0 ⋅ cosΘ)/0

This simplifies to:

v_0 ⋅ sinΘ - 10 = 50 - v_0 ⋅ cosΘ

Now we can solve for Θ:

v_0 ⋅ sinΘ = 50 + v_0 ⋅ cosΘ
tanΘ = (50 + v_0 ⋅ cosΘ)/v_0
Θ = arctan[(50 + v_0 ⋅ cosΘ)/v_0]

This gives us the angle at which we need to aim the launcher to hit the platform 10 cm above the floor, 50 cm away. Keep in mind that this is just a theoretical calculation and may not be exactly accurate in real life due to factors like air resistance and friction. It is always important to conduct experiments to verify theoretical calculations.