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!

answer: option a.

step-by-step explanation:

you know that the quadrilateral opqr ( a four sided figure) is inscribed in circle. this means that all its vertices are on the circle.

then, by definition, the opposite angles of the quadrilateral are supplementary, this means that the sum of their measures is 180 degrees.

you can observe in the figure that the angles ∠opq   and the angle ∠orq   are opposite, therefore:

then, can be used to d to calculate the measure of ∠opq