Question

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/s2. There are two equations that can be used to describe its motion over time:

x = x_0 + v_0t + 1/2at^2
v = v_0 + at

Would the balloon hit the ground before or after 1.0 s of falling? Which equation did you use to decide, and whal comparison did you make to determine that it would or would not hit the ground by then? Use 3-5 sentences to explain your reasoning and show all work including givens, unknowns, equation, substitutions and final answer

Answer 1

Givens: height x0 = 10.0 m, v0 = 0 (released from rest), a = 9.8 m/s^2. Unknown: does it reach the ground by t = 1.0 s? Use x = x0 + v0t + (1/2)at^2 and take the distance fallen in 1.0 s: d = (1/2)·9.8·(1.0)^2 = 4.9 m, which is less than 10.0 m, so it has not hit the ground by 1.0 s (it hits after 1.0 s). Solving for the impact time: (1/2)·9.8·t^2 = 10 → t = sqrt(20/9.8) ≈ 1.43 s, so the balloon hits the ground at about 1.43 s.