A major leaguer hits a baseball so that it leaves the bat at a speed of 30.5m/s and at an angle of 36.9 degrees above the horizontal. You can ignore air resistance.At what two times is the baseball at a height of 8.50 above the point at which it left the bat? What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat? What is the direction of the baseball's velocity when it returns to the level at which it left the bat?
physics - drwls, Sunday, September 21, 2008 at 11:40pm
The equation for height y (above the bat) vs time is
y = 30.5 sin 36.0 t - (1/2) g t^2
Set that equal to 8.5 m and solve for t.
The second question can be answered easily using energy considerations. The kinetic energy will be the same both times y = 0, since the potential energy will be the same.
The horizontal component of V will be the same, and the Vertical component will change direction but have the same magnitude.