A student drops a ball from a window 3.5 m above the sidewalk. The ball accelerates at 9.80 m/s squared. How fast is it moving when it hits the sidewalk?
I have no idea even how to start this problem, and if you figure it out could you please tell me how you got it and what type of formula you used. Thanks!
2007-12-19 11:07:40 · 4 answers · asked by UMP Skills 3 in Science & Mathematics ➔ Physics
I Think it is 35.3 becuz i used a calculator
hehe
2007-12-19 11:15:36 · answer #1 · answered by *Kathy* 2 · 0⤊ 1⤋
Use the formula for free-fall:
vf ^2 = vi^2 + 2gd
where vf = final velocity, vi = initial velocity, g is acceleration due to gravity and d is the distance.
Since the ball is dropped, the initial velocity is 0.
vf^2 = 0 + 2(9.8)(3.5)
vf^2 = 68.6
vf = 8.29 m/s
2007-12-19 11:16:09 · answer #2 · answered by Luke C 3 · 0⤊ 0⤋
v^2 - v0^2 = 2as
v^2 - 0 = (2)(9.80)(3.5)
v^2 = 68.6 m^2/s^2
v â 8.282512 m/s â 8.28 m/s
2007-12-19 11:22:33 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
3.5m
After first m = 9.8 m/s
After 2nd m = 19.6 m/s
After 3rd m = 30.4 m/s
At contact = 35.3 m/s
2007-12-19 11:13:33 · answer #4 · answered by edward_otto@sbcglobal.net 5 · 0⤊ 0⤋