For a, the y component of the vector is found in [tex] V_{y} = v_{0} sin \beta [/tex]. Our initial velocity is 70 and our angle is 22.5 so the y component of the vector is 27 m/s. For b, the x component is found the same way except we use the cos of the angle instead to get that the x component of the velocity vector is 65 m/s. I have to skip c and look into that one a bit more...for d. how high the ball travels is the equivalency of asking at what point in the air did it stop for a nanosecond and then turn around and head back down? It is at this point that the velocity is 0. So we use our velocity and displacement equation [tex] v^{2} = v_{0} ^{2}+2a*delta x [/tex]. Fill in the velocity found in the y component (cuz y is up and down movement and this is what they are wanting when they ask how high the object reached), a is gravity and the displacement is what we are looking for:[tex]0=729-19.6deltax[/tex]. Solve for delta x to get a height of 37.2 m. I'm assuming that the object in part e hits the ground which is below its initial point of trajectory, so I am going to use the range equation here but this time the initial velocity will use the x component cuz horizontal movement occurs along the x-axis. [tex]range= \frac{ v_{0} ^{2} sin(2 \beta )}{g} [/tex]. Filling that in you get that the range is 304.8 m. This parabolic stuff is the hardest part of the whole entire course of Physics. I still need to look into c...
