Standing on a bridge, you throw a 3.0 g stone straight upward with speed of 2.5 m/s. How long does it take to return to the original position?
while rising,
v = v0 + at
0 = 2.5 - 9.8t
t = .255
That's the time to stop rising.
Total time in the air is thus 0.510 sec
To find out how long it takes for the stone to return to the original position, we can use the laws of motion, specifically the equation for time. Here's how you can calculate it:
1. Identify the relevant information given in the question:
- Initial speed of the stone (u) = 2.5 m/s (upward)
- Acceleration due to gravity (g) = -9.8 m/s² (downward)
2. Determine the initial velocity in terms of positive and negative directions:
Since the stone is thrown upward, the initial velocity is positive (+2.5 m/s).
3. Use the appropriate equation of motion to find the time the stone takes to reach the highest point (where the velocity becomes zero):
The equation we will use is: v = u + at, where:
- v = final velocity (0 m/s at the highest point)
- u = initial velocity (2.5 m/s)
- a = acceleration due to gravity (-9.8 m/s²)
- t = time (to be calculated)
Plugging in the values, we can rearrange the equation to solve for t:
0 = 2.5 - 9.8t
4. Solve for t:
-9.8t = -2.5
t = -2.5 / -9.8
t ≈ 0.255 seconds
5. Since the stone takes the same amount of time to return to the original position as it does to reach the highest point, the total time is twice the time calculated in Step 4:
Total time = 2 * 0.255 seconds
Total time ≈ 0.51 seconds
Therefore, it takes approximately 0.51 seconds for the stone to return to its original position.