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.
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 reach the pavement on its initial drop, the time it takes to reach its maximum height during the bounce, and the time it takes to reach the boy's hand.
Step 1: Calculate the time for the ball to reach the pavement on the initial drop.
Using the equation:
s = ut + (1/2)at^2
where s is the distance, u is the initial velocity, a is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
We assume the initial velocity is zero, so the equation simplifies to:
s = (1/2)at^2
Plugging in the values:
s = 8.30 m
a = -9.8 m/s^2
Rearranging the equation to solve for t:
t = sqrt(2s/a)
t = sqrt(2 * 8.30 / -9.8) = 1.45 seconds
Step 2: Calculate the time for the ball to reach its maximum height during the bounce.
Since the ball rises to a height of 5.00 m, we can use the same equation:
s = (1/2)at^2
Plugging in the values:
s = 5.00 m
a = -9.8 m/s^2
Rearranging the equation to solve for t:
t = sqrt(2s/a)
t = sqrt(2 * 5.00 / -9.8) = 1.02 seconds
Note: The total time for the bounce is double this time since it takes the same amount of time to rise and fall during the bounce.
Step 3: Calculate the time for the ball to reach the boy's hand.
To calculate this time, we first need to find the total time that the ball spends in the air during the bounce. Since it takes 1.02 seconds for the ball to reach its maximum height, and the total time is double this, the ball spends 2.04 seconds in the air during the bounce.
To calculate the time for the ball to reach the boy's hand, we subtract the time spent during the bounce from the total time for the ball to reach the pavement:
Total time = time to reach the pavement + time during the bounce + time to reach the boy's hand
Total time = 1.45 s + 2.04 s + t
Using the values we calculated, we can substitute them into the equation and solve for t:
t = Total time - (1.45 s + 2.04 s)
t = Total time - 3.49 s
Since the ball reaches the boy's hand when it is 1.10 m above the pavement, we can use the equation for free fall distance to calculate the time it takes for the ball to reach this height:
s = ut + (1/2)at^2
Plugging in the values:
s = 1.10 m
u = 0 m/s
a = -9.8 m/s^2
Rearranging the equation to solve for t:
t = sqrt(2s/a)
t = sqrt(2 * 1.10 / -9.8) = 0.15 seconds
Finally, we substitute this value into the equation for the total time:
Total time = 1.45 s + 2.04 s + 0.15 s
Total time = 3.64 seconds
Therefore, the total amount of time that the ball is in the air, from drop to catch, is 3.64 seconds.
To find 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, the time it takes for the ball to rise after bouncing, and the time it takes for the ball to fall again until it is caught.
Let's start by calculating the time it takes for the ball to fall from the initial height of 8.30 m. We can use the equation:
h = (1/2) * g * t^2
where h is the initial height, g is the acceleration due to gravity, and t is the time.
Rearranging the equation, we get:
t = sqrt((2 * h) / g)
Plugging in the values, we have:
t = sqrt((2 * 8.30 m) / 9.8 m/s^2)
t = sqrt(16.6 / 9.8)
t = sqrt(1.69)
t = 1.30 s (rounded to two decimal places)
So, it takes approximately 1.30 seconds for the ball to fall from the initial height.
Next, let's calculate the time it takes for the ball to rise after bouncing. Since the ball rises to a height of 5.00 m, we can use the same equation:
t = sqrt((2 * h) / g)
Plugging in the values, we have:
t = sqrt((2 * 5.00 m) / 9.8 m/s^2)
t = sqrt(10.00 / 9.8)
t = sqrt(1.02)
t = 1.01 s (rounded to two decimal places)
Therefore, it takes approximately 1.01 seconds for the ball to rise after bouncing.
Finally, let's calculate the time it takes for the ball to fall again until it is caught. The distance the ball needs to fall is the height from the catch point (1.10 m) to the pavement (0 m), so we use the same equation:
t = sqrt((2 * h) / g)
Plugging in the values, we have:
t = sqrt((2 * 1.10 m) / 9.8 m/s^2)
t = sqrt(2.20 / 9.8)
t = sqrt(0.22)
t = 0.47 s (rounded to two decimal places)
Therefore, it takes approximately 0.47 seconds for the ball to fall again until it is caught.
To get the total time, we add up the individual times:
Total time = time to fall + time to rise + time to fall again
Total time = 1.30 s + 1.01 s + 0.47 s
Total time = 2.78 s (rounded to two decimal places)
Hence, the total amount of time that the ball is in the air, from drop to catch, is approximately 2.78 seconds.