Asked by Anonymous
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.
Answers
Answered by
Damon
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
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
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