A 3.61 kg ball is dropped from the roof of a building 152.5 m high. While the ball is falling to Earth, a horizontal wind exerts a constant force of 11.6 N on the ball. How long does it take to hit the ground? The acceleration of gravity is 9.81 m/s^2
hf=hi+vi*t-4.9t^2
Vi=0
hf=0,hi=152.5m, solve for time.
To find out how long it takes for the ball to hit the ground, we can use the equation of motion:
\(d = ut + \frac{1}{2}at^2\)
Where:
- \(d\) is the distance traveled (152.5 m in this case)
- \(u\) is the initial velocity (0 m/s since the ball is dropped)
- \(a\) is the acceleration (9.81 m/s^2 downward due to gravity)
- \(t\) is the time we're trying to find
Since the horizontal wind exerts a constant force on the ball, it doesn't affect the vertical acceleration due to gravity. Therefore, we only need to consider the vertical motion.
Applying the equation of motion to our problem, it becomes:
\(152.5 = 0 \times t + \frac{1}{2} \times 9.81 \times t^2\)
Rearranging the equation, we get:
\(4.905 t^2 = 152.5\)
Now, divide both sides of the equation by 4.905 to solve for \(t^2\):
\(t^2 = \frac{152.5}{4.905}\)
Calculating the right side of the equation, we have:
\(t^2 = 31.107\)
Taking the square root of both sides, we find:
\(t = \sqrt{31.107}\)
Evaluating the expression gives us:
\(t \approx 5.57\) seconds
Therefore, it takes approximately 5.57 seconds for the ball to hit the ground.