A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin returns to the juggler's hand
time to the top is when
v = 8.20 - 9.81t = 0
find t, then double that, since the pin has to fall back to the ground.
To find the time it takes for the bowling pin to return to the juggler's hand, we need to consider the motion of the pin.
When the pin is thrown straight up, it will experience a vertical motion influenced by gravity. The key here is to realize that the time it takes for the pin to reach its maximum height is the same as the time it takes to fall back to the juggler's hand. This is because the pin follows a symmetric trajectory.
To find the time it takes for the pin to reach its maximum height, we can use the equation:
v = u + at
Where:
v is the final velocity (0 m/s as the pin momentarily comes to rest at its highest point),
u is the initial velocity (8.20 m/s as the pin is thrown straight up),
a is the acceleration due to gravity (-9.8 m/s², considering gravity pulls the pin downward).
Let's rearrange the equation to solve for time (t):
t = (v - u) / a
Since the final velocity (v) is equal to zero, we have:
t = -u / a
Substituting the given values:
u = 8.20 m/s
a = -9.8 m/s²
t = -8.20 m/s / (-9.8 m/s²)
Calculating this expression gives us:
t ≈ 0.837 seconds
Therefore, it takes approximately 0.837 seconds for the bowling pin to return to the juggler's hand after being thrown straight up.