from the top of a 35.3-m tall building. The ball then passes the top of a window that is 13.3 m above the ground 2.00 s after being thrown. What is the speed of the ball as it passes the top of the window?
2007-09-06 04:58:16 · 2 answers · asked by Fashionably lame 1 in Science & Mathematics ➔ Physics
Let the distance from the top of the building to the top of the window = d = 35.5 - 13.3
Now d = -1/2 gt^2 - v0t g = 9.8 m/s^2 and t = 2 sec.
re-arrange --> v0 = (d - 1/2gt^2)/t = (22.5 - 1/2*9.8*4)/2 =1.45 m/s
The speed at window v = -gt - v0 = -21.05 m/s with the minus sign indicating the ball is going down.
2007-09-06 05:28:56 · answer #1 · answered by nyphdinmd 7 · 0⤊ 1⤋
according to energy conservation
mgH=mgh+1/2mv^2
g*35.3=g*13.3+1/2*v^2
g*22=1/2v^2
v=(440)^1/2
v=20.9m/sec
2007-09-06 05:27:35 · answer #2 · answered by miinii 3 · 0⤊ 0⤋