Anonymous
Jul 8
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
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
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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