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A student holds a water balloon outside of an open window and lets go. The window is 10 meters above the ground, and the balloon is falling under the acceleration of gravity, which is 9.8 m/s^2. There are two equations that can be used to describe its motion over time:
x= x0 + v0t + 1/2 at^2
v= v0 + at
Would the balloon hit the ground before or after 1.0 s of falling? Which equation did you use to decide, and what comparison did you make to determine that it would or would not hit the ground by then?
Use 3 sentences.
oobleck
answered
2 years ago
2 years ago
Evee
answered
2 years ago
2 years ago
I did yf = y0 + bot + 1/2at^2 (1)
So 0 = 10m 1/2gt^2
Then t = 2 + 10m/9.8/8^2 = 1.4s.
Evee
answered
2 years ago
2 years ago
Not bot but *vot*
Evee
answered
2 years ago
2 years ago
We add on to 4.9 meters?
Anonymous
answered
2 years ago
2 years ago
0 = 10 + 0 - (1/2) (9.81) t^2
4.9 t^2 = 10
t = 1.43 seconds in the air before hitting ground
so
it is still in the air after one second, like period :)
Anonymous
answered
2 years ago
2 years ago
and remember if h is up, g is down, negative
Evee
answered
2 years ago
2 years ago
Thank you :)
DrBob222
answered
2 years ago
2 years ago
I'm still trying to figure out how biology plays into this question.
Copypasteking
answered
2 years ago
2 years ago
Thx
Brandon
answered
2 years ago
2 years ago
Is anyone still on this cuz I need help still
Jordan
answered
2 years ago
2 years ago