Little help with Projectile Motion problem

[sun2k4]

Homework Statement

Mike is playing dart at home. Suppose he releases the dart at 15.0 m/s. His releasing
position is 5cm above the dart board center. The horizontal distance between the release position
and the board is 2.5m. How should Mike aim, i.e., what are the angle α and Yoffset as shown on the
graph below? Keep three significant digits, and you need to use the small angle approximation of
cosα=1 when applicable. This approximation is needed once, and only once.

Homework Equations

Range = (V^2/g) sin 2x

The Attempt at a Solution

I tried to use the above equation to solve for sin2x and was able to, but I cannot figure out how calculate the angle without assuming a parabola (the above equation does assume that) because there is a 5cm Y difference between launching point and the point where the dart hits the target, could anyone try to guide me in the right direction here? Any input would be really appreciated

 

[gneill]

The range equation that you quote is only applicable when the starting height is equal to the landing height. This is not the case here.

I'm afraid you'll have to write the equations of motion and "do the math".

 

[sun2k4]

Hi gneill, thank you very much for your answer, could you explain a little bit more? would you be referring to the position functions in terms of time?

 

[gneill]

Hi gneill, thank you very much for your answer, could you explain a little bit more? would you be referring to the position functions in terms of time?

Yes. The "usual" projectile motion equations.

 

[rwisz]

.Hi there! It looks like you're on the right track with using the range equation to solve for the angle α. However, since there is a 5cm difference in the vertical position of the dart, we need to take that into account in our calculations.

First, let's define some variables:
V = 15.0 m/s (initial velocity)
g = 9.8 m/s^2 (acceleration due to gravity)
x = 2.5 m (horizontal distance between release position and target)
y = 5 cm (vertical distance between release position and target)

Using the range equation, we can write:
x = (V^2/g) * sin 2α

Now, we need to take into account the 5 cm difference in the vertical position. This means that at the horizontal distance x, the dart will actually be 5 cm below the target. So, we can write:
x = (V^2/g) * sin 2α - y

Substituting in our values, we get:
2.5 = (15.0^2/9.8) * sin 2α - 0.05

Solving for sin 2α, we get:
sin 2α = (2.5 + 0.05) * 9.8 / 15.0^2 = 0.00131

Now, using the small angle approximation of cos α = 1, we can write:
sin 2α ≈ 2α
So, we have:
2α = 0.00131
α = 0.000655 radians

Converting to degrees, we get:
α ≈ 0.0375 degrees

So, Mike should aim at a very small angle of approximately 0.0375 degrees and he should also offset his aim by 5 cm below the target. I hope this helps! Let me know if you have any further questions.