To solve the problem of the water balloon falling from a height of 10 meters, we need to use the correct kinematic equation for motion under constant acceleration. The two equations that are commonly used are:
- \( d = v_0 t + \frac{1}{2} a t^2 \)
- \( v = v_0 + at \)
Where:
- \( d \) is the distance fallen,
- \( v_0 \) is the initial velocity (which is 0 m/s as the balloon is simply dropped),
- \( a \) is the acceleration due to gravity (9.8 m/s²),
- \( t \) is the time in seconds.
Question 1:
To determine whether the balloon would hit the ground before or after 1.0 seconds of falling, we'll use the first equation: \[ d = v_0 t + \frac{1}{2} a t^2 \]
Substituting the values:
- \( v_0 = 0 \) m/s,
- \( a = 9.8 \) m/s²,
- \( t = 1.0 \) s,
We have: \[ d = 0 + \frac{1}{2} (9.8 , \text{m/s}^2)(1.0 , \text{s})^2 = 4.9 , \text{m} \]
After 1 second, the balloon has fallen 4.9 meters. Since the window is 10 meters above the ground, the balloon is still 5.1 meters above the ground after 1 second. Therefore, the balloon will not hit the ground before 1 second.
Answer for Question 1:
The correct equation is #1. It would not hit the ground before 1 second.
Question 2:
For this part, we are asked for the distance fallen in 1 second using the correct equation. The equation to use is:
\[ X = v_0 t + \frac{1}{2} a t^2 \]
Substituting the correct values as previously calculated (with \( v_0 = 0 \) m/s):
For 1 second, we found: \[ X = (0 , \text{m}) + (0 , \text{m/s})(1.0 , \text{s}) + \frac{1}{2} (9.8 , \text{m/s}^2)(1, \text{s})^2 = 4.9, \text{m} \]
Answer for Question 2:
X = (0 m) + (0 m/s)(1.0 s) + ½ (9.8 m/s²)(1s)² = 4.9 m.