A baseball is thrown from the roof of a 24.0-m-tall building with an initial velocity of magnitude 10.0 m/s and directed at an angle of 53.1° above the horizontal.

Answer 1

oh well, I suppose I can pretend to know what the question is.

v = vertical speed
Vi = initial vertical speed
so
Vi = 10 sin 53.1

How long rising?
v = Vi - g t = Vi - 9.81 t
at top, v = 0
so
tt = Vi/9.81 at the top (calling tt the top time)

How high at top?
h = 24 + Vi tt - 4.9 tt^2

when at ground? (using tg)
0 = 24 + Vi tg - 4.9 tg^2
solve quadratic for tg at ground

how far?
constant u = 10 cos 53.1
d = u * tg