To find the height of an overhead power line, you throw a ball straight upward. The ball passes the line on the way up after 0.65 s, and passes it again on the way down 1.9 s after it was tossed.
(a) What is the height of the power line? (m)
(b) What is the initial speed of the ball? (m/s)
To find the height of the overhead power line and the initial speed of the ball, we can use the equations of motion for vertically thrown objects.
Let's start with part (a), finding the height of the power line:
1. First, we need to find the time it took for the ball to reach its highest point. Since the ball passed the power line on the way up after 0.65 s, the time to the highest point would be half of that, which is 0.65 s / 2 = 0.325 s.
2. Next, we can use the formula for the height at a given time for an object in free fall: h(t) = v₀t - (1/2)gt², where h is the height, v₀ is the initial velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s²).
With the known values from the problem, our equation becomes:
h(0.325 s) = v₀(0.325 s) - (1/2)(9.8 m/s²)(0.325 s)²
3. Rearranging the equation, we get:
h(0.325 s) = (v₀(0.325 s)) - (0.164 m/s²)(0.105625 s²)
h(0.325 s) = 0.105625v₀ - 0.001737359375 m
Now, we can use the information that the ball passes the power line again on the way down 1.9 s after it was tossed.
4. At this point, the total time would be 1.9 s + 0.325 s = 2.225 s.
Again, using the equation for height at a given time, we have:
h(2.225 s) = 0.105625v₀ - 0.001737359375 m
5. Since the height of the power line remains the same, we can set the two equations equal to each other:
0.105625v₀ - 0.001737359375 m = 0.105625v₀ - 0.001737359375 m
The units on both sides cancel, leaving us with:
0.105625v₀ = 0.105625v₀
We see that the equation is satisfied for any value of v₀, meaning that the height does not depend on the initial velocity.
6. Therefore, to find the height of the power line, we can choose either equation. Let's use the first equation:
h(0.325 s) = 0.105625v₀ - 0.001737359375 m
Plugging in the values:
0.001737359375 m = 0.105625v₀ - 0.001737359375 m
Simplifying the equation, we get:
2 * 0.001737359375 m = 0.105625v₀
0.00347471875 m = 0.105625v₀
v₀ = 0.00347471875 m / 0.105625
7. Calculating v₀:
v₀ ≈ 0.033 m/s
So, the height of the power line is 0.00347471875 m, and the initial speed of the ball is approximately 0.033 m/s.
Note: In this problem, the height of the overhead power line is given relative to the initial position of the thrown ball. If a reference point for the ground level is provided, the height can be calculated accordingly.