Need some Help please PHYSICS problem :)?

Question

A baseball is seen to pass upward by a window 29 m above the street with a vertical speed of 8 m/s. The ball was thrown from the street.

(a) What was its initial speed?...m/s
(b) What altitude does it reach?....m
(c) How long after it was thrown did it pass the window?....s
(d) After how many more seconds does it reach the street again? ....s

please also explain to me how to do it, i would really appreaciate it ..Thanks :)

Answer 1

(a) Use the rule v1*2 - v0^2 = 2ax1 (from the ref.), and thus
v0 = sqrt(v1^2 - 2ax1) with v1 = 8 m/s, a = -9.81 m/s^2 and x1 = 29 m. This gives V0 = 25.159 m/s.
(b) Now use xmax=at^2/2 = v0t/2 = v0^2/(2a) = 32.261 m.
(c) Next, average speed v (for the trip from ground to window) = (v1 + v0)/2, and time t = xmax/v = 2xmax/(v1 + v0) = 1.749 s.
(d) Use xmax=at^2/2 ==> t=sqrt(2xmax/a) = 2.565 s. That's the time to reach xmax and also the time to fall from xmax to the ground. So the time to reach the ground after passing the window on the way up = 2*2.565-1.749 s.

Answer 2

Kirchwey got it right, and I gave him a 'thumbs up' for it, but (to me) my way is more intuitive:

Since it passed the window at 8 m/s, the formula y = V²/2g will give the distance it goes above the window:
Y1 = 8²/(2*9.8) = 3.265 m.......add this to the 29 m height of the window to get Ymax = 32.265 m

This value of Ymax can be used in the same formula solved for V to find Vo = √(2gYmax) = 25.147 m/s

The time from passing the window to Ymax is t = Vw/g = 8/9.8 = .816 sec.....subtracted from the time to rise to Ymax = Vo/g - .816 = 2.566 - .816 = 1.75 sec. to get up to the window. t added to the time to fall from Ymax = T = 25.147/9.8 = 2.566 → .816 + 2.566 = 3.382 sec. (the time from 1st window passage to hitting the street again).