Suppose that a projectile is launched with a velocity of vo = 3.49±0.02 (m/s) at an angle of θ = 76◦ ±1 ◦ . after traveling a horizontal distance of x = 0.5 m, what is the uncertainty δx?
apply
δx = x/vo cosθ√{(cosθ) 2 (δvo) 2 +(vo sinθ) 2 (δθ) 2}

remember, the uncertainty of the angle must be expressed in radians to obtain the correct result.

δx=0.035 m

Step-by-step explanation:

We use the expression of the uncertainty of x:

δx = (x/(vo cosθ))*sqrt((cosθ)² (δV₀)²+(V₀ sinθ)²(δθ)² )

First of all, we need to identify δV₀ and δθ. In our case  δV₀ = 0.002 m/s and δθ = 0.0175 rad.

Remember: 1° = 0.0175 rad. This is very important to get the right answer here.

Now, we just need to replace the values we have into the equation δx:

δx = (0.5/(3.49*cos(76)))*sqrt((cos(76))²*(0.02)²+(3.49*sin(76))²*(0.0175)²)

δx  = 0.035 m.

Have a nice day!

17

Step-by-step explanation:

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