Baseball physics!!?

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 38.0 m/s and at an angle of 36.9° above the horizontal. You can ignore air resistance. (Assume upward is positive.)

a)At what two times is the baseball at a height of 13.0 m above the point at which it left the bat?
first time ________s
second time_______s

b) Calculate the horizontal and vertical components of the baseball's velocity at each of the two times calculated in part (a).
first time
____m/s (horizontal)
____m/s (verticle)
Second Time
____m/s (horizontal)
_____m/s (verticle)
(c) What is the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat?
_____m/s
______ degree (below the horizontal)

Answer 1

In the vertical above the launch point, with no air resistance, the ball will follow the following equation
y(t)=38*sin(36.9)*t-.5*g*t^2
plug 13 in for y(t) and solve for t. There will be two roots, which are the answers for a.

b)
The horizontal will be constant at
vx=38*cos(36.9)
the vertical will follow the equation
vy(t)=38*sin(36.9)-g*t
plug in the two times calculated above and solve for vy

c)because there is no air resistance, the flight is parabolic and symmetric about the midpoint, which is also the apogee of flight.
the speed will be 38.0 m/s at and angle of 36.9 below the horizontal.

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