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.

User Icon for oobleck oobleck answered
2 years ago

in 1 second, it falls 4.9 meters (1/2 at^2)

User Icon for Evee Evee answered
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.

User Icon for Evee Evee answered
2 years ago

Not bot but *vot*

User Icon for Evee Evee answered
2 years ago

We add on to 4.9 meters?

User Icon for Anonymous Anonymous answered
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 :)

User Icon for Anonymous Anonymous answered
2 years ago

and remember if h is up, g is down, negative

User Icon for Evee Evee answered
2 years ago

Thank you :)

User Icon for DrBob222 DrBob222 answered
2 years ago

I'm still trying to figure out how biology plays into this question.

User Icon for Copypasteking Copypasteking answered
2 years ago

Thx

User Icon for Brandon Brandon answered
2 years ago

Is anyone still on this cuz I need help still

User Icon for Jordan Jordan answered
2 years ago

Y’all I need help