posted by peter on .
A baseball is seen to pass upward by a window 27 m above the street with a vertical speed of 12 m/s. The ball was thrown from the street.
(a) What was its initial speed?
(b) What altitude does it reach?
(c) How long after it was thrown did it pass the window?
(d) After how many more seconds does it reach the street again?
why do communities need laws?
x = 1/2 a t^2 + v0 t + x0
and v = at + v0
(a)needs no translation, but some work: with v=12m/s, solve the velocity equation for the special time t (call it t) when the ball is at the window (in terms of 12m/sec and v0). Don't lose your nerve. Now plug that expression for the t1 into the x equation, with x equal to 27 m, and solve for v0.
(b) find the time, t2, when v=0 (That's at the top of the path.) plug that time into the position equation to find the x value at the highest point.
(c)Now that you know v0, return to the velocity equation in part a and solve for t.
(d)Use the position equation and ask "what time is it when x = 0?"