A baseball hit just above the ground leaves the bat at 22 m/s, at 53.6° above the horizontal. What is the maximum height of the ball above the ground?
Help!!!
u = 22 cos 53.6 forever
Vi = 22 sin 53.6
v = Vi - 9.8 t
at top v = 0
0 = 22 sin 53.6 - 9.8 t
solve for t, time to top
h = 0 + Vi t - 4.9 t^2