Question

A baseball is thrown at a 22.5 degree angle with an initial velocity of 70 m/s. Assume no air resistance and remember that the acceleration due to gravity is -9.8m/s2
a. what is the initial vertical component of the ball's velocity?
b.what is the horizontal component of the ball's velocity?
c.how long until the ball hits the ground?
d.how high did the ball travel?
e.how far did the ball travel horizontally when it hit the ground?

Must show all work. PLEASE HELP ME WITH THIS

Answer 1

For a, the y component of the vector is found in $V_{y} = v_{0} sin \beta$.  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 $v^{2} = v_{0} ^{2}+2a*delta x$.  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:$0=729-19.6deltax$.  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. $range= \frac{ v_{0} ^{2} sin(2 \beta )}{g}$.  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...