A golf ball is dropped from rest from a height of 8.30 m. It hits the pavement, then bounces back up, rising just 5.00 m before falling back down again. A boy then catches the ball when it is 1.10 m above the pavement. Ignoring air resistance, calculate the total amount of time that the ball is in the air, from drop to catch.
https://answers.yahoo.com/question/index?qid=20120830224254AAQzEep
a very similar problem.
To calculate the total amount of time that the ball is in the air, we need to calculate the time it takes for the ball to fall from the initial height to the pavement, the time it takes for the ball to rise from the pavement to its maximum height, and finally, the time it takes for the ball to fall from its maximum height to the catch point.
We can use the kinematic equation for free-falling objects:
h = (1/2) * g * t^2,
where h represents the height, g represents the acceleration due to gravity (9.81 m/s^2), and t represents time.
First, let's calculate the time it takes for the ball to fall from the initial height of 8.30 m to the pavement:
h = (1/2) * g * t^2
8.30 m = (1/2) * 9.81 m/s^2 * t^2
Simplifying:
t^2 = (2 * 8.30 m) / 9.81 m/s^2
t^2 = 1.694 s^2
t = √(1.694 s^2)
t ≈ 1.30 s
Now, let's calculate the time it takes for the ball to rise from the pavement to its maximum height of 5.00 m:
Since the ball is thrown vertically up, we can use the equation:
v = u + g * t,
where v represents the final velocity, u represents the initial velocity (which is equal to the final velocity when the ball hits the pavement), g represents the acceleration due to gravity, and t represents time.
The final velocity when the ball hits the pavement is given by:
v = u + g * t,
0 m/s = u + 9.81 m/s^2 * 1.30 s,
Simplifying:
u = -9.81 m/s * 1.30 s,
u ≈ -12.73 m/s (where negative sign indicates the velocity is in the opposite direction)
Using the equation:
v = u + g * t,
0 m/s = -12.73 m/s + 9.81 m/s^2 * t,
Simplifying:
t = (12.73 m/s) / 9.81 m/s^2,
t ≈ 1.30 s
Now, let's calculate the time it takes for the ball to fall from its maximum height of 5.00 m to the catch point at 1.10 m above the pavement:
Using the same kinematic equation:
h = (1/2) * g * t^2
5.00 m - 1.10 m (height difference) = (1/2) * 9.81 m/s^2 * t^2
Simplifying:
3.90 m = 4.905 m/s^2 * t^2
t^2 = (3.90 m) / 4.905 m/s^2
t^2 ≈ 0.795 s^2
t ≈ √(0.795 s^2)
t ≈ 0.89 s
Finally, to find the total time of flight, we add the individual times together:
Total time = time to fall + time to rise + time to fall again
Total time = 1.30 s + 1.30 s + 0.89 s
Total time ≈ 3.49 s
Therefore, the total amount of time that the ball is in the air, from drop to catch, is approximately 3.49 seconds.