I asked this question earlier and the response seems correct, but I cannot figur

Question

I asked this question earlier and the response seems correct, but I cannot figure out how they got 11.18 theta. My trig is not so strong, albeit I know the basic principles (sohcahtoa), the 11.18 confuses me. I'll award points to someone who can break it down just a bit more for me. The question was:

                        An individual threw an object at a clock that was 5 meters away. The clock was 120cm. above the point from where they initially threw the object. The velocity that was obtained was 25 m/s. What are the horizontal and vertical of the initial velocity?

Their response was:

                                         h=Vi^2*sin(theta)^2/2*g

                                         theta=11.18

                                        Vhor=Vi*cos(theta)=24.52m/s

                                         Vvert=Vi*sin(theta)=4.85 m/s

Explanation / Answer

h=Vi^2*sin(theta)^2/2*g

(sin(theta))^2 = h*2*g / Vi^2

sin (theta) = sqrt (h*2*g / Vi^2)

theta = sin^-1 (sqrt(h*2*g / Vi^2)) (sin^-1 = Inverse of sin)

this way putting values theta can be got.

theta = 11.18

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