posted by John on .
A major leaguer hits a baseball so that it leaves the bat at a speed of 31.5 and at an angle of 36.1above the horizontal. You can ignore air resistance.
At what two times is the baseball at a height of 8.00 above the point at which it left the bat?
You need to give the dimensions of speed and height. I will assume they are m/s and meters
Solve the equation
Y = 8.00 = 31.5 t sin36.1 - 4.90 t^2
8.00 = 18.56 t - 4.90 t^2
The quadratic equation will provide two answers