The motion of the water balloon can be described by the equations of motion under constant acceleration. In this case, the acceleration is due to gravity (approximately \(9.8 , \text{m/s}^2\) downward). Given that the balloon is released from rest, the initial velocity (\(v_0\)) is \(0 , \text{m/s}\).
1. Velocity Equation
You provided the velocity equation:
\[ v = v_0 + at \]
Substituting the initial conditions:
- \(v_0 = 0 , \text{m/s}\) (the initial velocity)
- \(a = 9.8 , \text{m/s}^2\)
The equation simplifies to:
\[ v = 0 + (9.8 , \text{m/s}^2)(t) = 9.8t \]
2. Displacement Equation
The displacement (\(s\)) of the balloon from the point of release can be represented by the equation:
\[ s = v_0 t + \frac{1}{2} a t^2 \]
Again substituting the initial conditions (\(v_0 = 0\)):
\[ s = 0 + \frac{1}{2}(9.8 , \text{m/s}^2)t^2 = 4.9t^2 \]
Total Displacement from the Window to the Ground
Given that the window is \(10 , \text{m}\) above the ground, the total displacement equation from the point of release until it hits the ground would be:
\[ 10 = 4.9t^2 \]
Find Time (\(t\))
Rearranging the equation to solve for \(t\):
\[ t^2 = \frac{10}{4.9} \]
Calculating this:
\[ t^2 \approx 2.04 \] \[ t \approx \sqrt{2.04} \approx 1.43 , \text{s} \]
Conclusion
The water balloon takes approximately \(1.43\) seconds to reach the ground from a height of \(10\) meters. You can also find the final velocity just before it hits the ground by substituting \(t\) back into the velocity equation:
\[ v = 9.8t = 9.8(1.43) \approx 14.0 , \text{m/s} \]
Therefore, the water balloon hits the ground at a speed of approximately \(14.0 , \text{m/s}\).
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