A baseball is thrown from the stands 32 ft above the field at an angle of 30 degrees up from the horizontal. When and how far away will the ball strike the ground if its initial speed is 32ft/sec?
My attempt: Vertical $r(t)$ is given as $$32\sin(30^\circ)t - 1/2(gt^2) = -32$$ then I get $$16t - (9.81/2)t^2 = -32$$ which is equivalent to:
$$16t-(9.81/2)t^2 = -32$$
Then do quadratic equation and get $t = 4.66$ sec.
However, the textbook describes g as 32ft/sec$^2$. and gives $t = 2$ and distance as 55.4 ft.
What am I doing wrong?