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.
0 = 10 + 0 - (1/2) (9.81) t^2
4.9 t^2 = 10
t = 1.43 seconds in the air before hitting ground
it is still in the air after one second, like period :)
in 1 second, it falls 4.9 meters (1/2 at^2)
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.