drop is vi = 0 m/s
How far is pf=?
Set up and solve... plz help
2006-09-17 17:10:40 · 5 answers · asked by sub.lihhjj 2 in Science & Mathematics ➔ Physics
S=ut + 1/2 at^2
u=0 t=1 a=9.81m/s^2 (gravity)
Thus S= 0 + 1/2x9.81x1x1
S=9.81/2 = 4.905m = 16.081 feet
2006-09-17 17:12:47 · answer #1 · answered by Tammi J 3 · 0⤊ 0⤋
Ignoring air resistance.
y=y0+Voyt+1/2gt^2
Yinitial = 0 initial velocity = 0 g = -9.8 m/s^2
- 1/2 (9.8 m/s^2) x 1 s = -4.9 m since it's distance, it should be 4.9 m = 16 ft. That's from the origin of releasing the ball.
2006-09-18 00:17:31 · answer #2 · answered by Omar 2 · 0⤊ 0⤋
It's to very easy
y = 1/2 a t^2
y = 4.9 (1s)^2 = 4.9
The other answer are wrong... Be carefully
2006-09-18 04:39:18 · answer #3 · answered by Juan D 3 · 0⤊ 1⤋
32 feet
2006-09-18 00:12:38 · answer #4 · answered by Rich Z 7 · 0⤊ 1⤋
depends on how tall you are...
jk its 16 feet...
2006-09-18 00:26:32 · answer #5 · answered by Anonymous · 0⤊ 0⤋