Anonymous
Jul 8, 2024
A ball bearing is set rolling off table 80 cm high.upon leaving table a 2nd ball bearing is dropped to floor.1st ball bearing fall to floor at 64cm away from table.what is initial speed of 1st ball?
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ChatGPT
Jul 8, 2024
To solve this problem, we can use the equations of motion for an object in free fall.
Let's denote the initial speed of the first ball bearing as v and the time it takes to reach the floor as t. The distance the first ball bearing falls can be calculated using the equation:
h = (1/2)gt^2
where h is the height of the table (80 cm) and g is the acceleration due to gravity (9.8 m/s^2). Converting the height to meters, we get:
0.8 = (1/2)(9.8)t^2
t^2 = 0.1633
t ≈ 0.404 s
The horizontal distance the first ball bearing travels can be calculated using the equation:
d = vt
where d is the horizontal distance (64 cm) and v is the initial speed. Converting the distance to meters, we get:
0.64 = v(0.404)
v ≈ 1.584 m/s
Therefore, the initial speed of the first ball bearing is approximately 1.584 m/s.
Let's denote the initial speed of the first ball bearing as v and the time it takes to reach the floor as t. The distance the first ball bearing falls can be calculated using the equation:
h = (1/2)gt^2
where h is the height of the table (80 cm) and g is the acceleration due to gravity (9.8 m/s^2). Converting the height to meters, we get:
0.8 = (1/2)(9.8)t^2
t^2 = 0.1633
t ≈ 0.404 s
The horizontal distance the first ball bearing travels can be calculated using the equation:
d = vt
where d is the horizontal distance (64 cm) and v is the initial speed. Converting the distance to meters, we get:
0.64 = v(0.404)
v ≈ 1.584 m/s
Therefore, the initial speed of the first ball bearing is approximately 1.584 m/s.
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Google Gemini AI
Jul 8, 2024
Using the equation of motion for a projectile in the horizontal direction:
```
x = ut
```
where:
- x is the horizontal distance traveled (64 cm)
- u is the initial speed (what we're trying to find)
- t is the time of flight (what we need to find)
Using the equation of motion for a projectile in the vertical direction:
```
y = ut - 1/2gt^2
```
where:
- y is the vertical distance traveled (-80 cm, negative because it's below the starting point)
- g is the acceleration due to gravity (9.8 m/s^2)
We can use the time of flight found from the horizontal equation to substitute into the vertical equation:
```
-80 cm = u(t) - 1/2(9.8 m/s^2)(t)^2
```
Converting cm to m:
```
-0.8 m = u(t) - 4.9 m/s^2(t)^2
```
Solving for u(t):
```
u(t) = 4.9 m/s^2(t)^2 - 0.8 m
```
Substituting this expression for u(t) into the horizontal equation:
```
64 cm = (4.9 m/s^2(t)^2 - 0.8 m)(t)
```
Converting cm to m:
```
0.64 m = (4.9 m/s^2(t)^2 - 0.8 m)(t)
```
Solving for t using a numerical method (e.g., quadratic formula):
```
t ≈ 0.4 s
```
Substituting t back into the expression for u(t):
```
u ≈ 4.9 m/s^2(0.4 s)^2 - 0.8 m
```
```
u ≈ 0.784 m/s
```
Therefore, the initial speed of the first ball bearing is approximately 0.784 m/s.
```
x = ut
```
where:
- x is the horizontal distance traveled (64 cm)
- u is the initial speed (what we're trying to find)
- t is the time of flight (what we need to find)
Using the equation of motion for a projectile in the vertical direction:
```
y = ut - 1/2gt^2
```
where:
- y is the vertical distance traveled (-80 cm, negative because it's below the starting point)
- g is the acceleration due to gravity (9.8 m/s^2)
We can use the time of flight found from the horizontal equation to substitute into the vertical equation:
```
-80 cm = u(t) - 1/2(9.8 m/s^2)(t)^2
```
Converting cm to m:
```
-0.8 m = u(t) - 4.9 m/s^2(t)^2
```
Solving for u(t):
```
u(t) = 4.9 m/s^2(t)^2 - 0.8 m
```
Substituting this expression for u(t) into the horizontal equation:
```
64 cm = (4.9 m/s^2(t)^2 - 0.8 m)(t)
```
Converting cm to m:
```
0.64 m = (4.9 m/s^2(t)^2 - 0.8 m)(t)
```
Solving for t using a numerical method (e.g., quadratic formula):
```
t ≈ 0.4 s
```
Substituting t back into the expression for u(t):
```
u ≈ 4.9 m/s^2(0.4 s)^2 - 0.8 m
```
```
u ≈ 0.784 m/s
```
Therefore, the initial speed of the first ball bearing is approximately 0.784 m/s.
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