1. 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 0.90 s, and passes it again on the way down 1.4 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
2.While riding on an elevator descending with a constant speed of 3.4 m/s, you accidentally drop a book from under your arm.
(a) How long does it take for the book to reach the elevator floor, 1.4 m below your arm?
s
(b) What is the book's speed relative to you when it hits the elevator floor?
m/s
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1. To find the height of the power line, we can use the kinematic equations of motion.
(a) First, we need to find the time it takes for the ball to reach its maximum height. Since the ball passes the line on the way up after 0.90 s, we can divide this time by 2 to find the time for the ascent: t_up = 0.90 s / 2 = 0.45 s.
Using the equation for displacement with constant acceleration:
Δy = v_i * t + (1/2) * a * t^2
For the upward motion, the final velocity is 0 m/s at the maximum height, so v_f = 0 m/s.
The initial velocity is unknown, denoted as v_i.
The acceleration due to gravity is -9.8 m/s^2, with negative sign indicating the opposite direction (upwards).
At the highest point, the displacement is equal to the height of the power line, h. Thus, we can write the equation:
h = v_i * t_up + (1/2) * (-9.8 m/s^2) * (0.45 s)^2
Solving for h, we get:
h = v_i * 0.45 s - 0.99 m
(b) Next, we need to find the initial velocity of the ball.
Since the ball passes the line on the way down 1.4 s after being tossed, we can subtract the time it takes for the ascent (0.45 s) from this total time to find the time for the descent: t_down = 1.4 s - 0.45 s = 0.95 s.
Using the equation for displacement with constant acceleration again, this time for the downward motion:
Δy = v_i * t + (1/2) * a * t^2
At the height of the power line, the displacement is 0. Thus, we can write the equation:
0 = v_i * t_down + (1/2) * (-9.8 m/s^2) * (0.95 s)^2
Solving for v_i, we get:
v_i = (0.5 * 9.8 m/s^2 * (0.95 s)^2) / 0.95 s
v_i = 4.9 m/s
Therefore, the height of the power line is 4.9 m and the initial speed of the ball is 4.9 m/s.
2. For this problem, we can use the kinematic equations of motion again.
(a) Since the elevator is descending with a constant speed, the book will also have a constant downward speed of 3.4 m/s. The distance from your arm to the elevator floor is 1.4 m.
Using the equation for displacement with constant velocity:
Δy = v * t
We can plug in the values:
1.4 m = 3.4 m/s * t
Solving for t, we get:
t = 1.4 m / 3.4 m/s
t ≈ 0.41 s
Therefore, it will take approximately 0.41 second for the book to reach the elevator floor.
(b) The velocity of the book relative to you will be the difference between the elevator's downward velocity (3.4 m/s) and the book's downward velocity (0 m/s since it was initially dropped and not thrown).
v_book_relative = velocity_of_elevator - velocity_of_book
v_book_relative = 3.4 m/s - 0 m/s
v_book_relative = 3.4 m/s
Therefore, the book's speed relative to you when it hits the elevator floor is 3.4 m/s.
To solve these problems, we can use the kinematic equations of motion. These equations describe the motion of objects under the influence of gravity.
1. (a) To find the height of the power line, we can use the equation for the position of an object in free fall:
h = (1/2) * g * t^2,
where h is the height, g is the acceleration due to gravity (around 9.8 m/s^2 on Earth), and t is the time.
In this case, we have two different times when the ball passes the power line. The time it takes on the way up is 0.90 s, and the time it takes on the way down is 1.4 s. We can set up two equations using these times:
h = (1/2) * g * (0.90 s)^2,
h = (1/2) * g * (1.4 s)^2.
Simplifying these equations, we get:
h = 0.405 g,
h = 0.98 g.
Since both equal h, we can set them equal to each other:
0.405 g = 0.98 g.
Solving for g, we find:
g = 0.405/0.98 = 0.4138 m.
Thus, the height of the power line is given by:
h = (1/2) * g * t^2 = (1/2) * 0.4138 m/s^2 * (0.90 s)^2 = 0.166 m.
(b) To find the initial speed of the ball, we can use the kinematic equation:
v = g * t,
where v is the initial speed, g is the acceleration due to gravity, and t is the time. We can use either of the two times given to calculate the initial speed. Let's choose the time it takes on the way up, which is 0.90 s:
v = g * t = 0.4138 m/s^2 * 0.90 s = 0.3725 m/s.
Thus, the initial speed of the ball is 0.3725 m/s.
2. (a) To find how long it takes for the book to reach the elevator floor, we need to determine the time it takes for the book to fall a distance of 1.4 m. We can use the kinematic equation:
h = (1/2) * g * t^2,
where h is the height, g is the acceleration due to gravity, and t is the time.
Rearranging the equation to solve for t, we have:
t = sqrt(2h / g).
Plugging in the values, we get:
t = sqrt(2 * 1.4 m / 9.8 m/s^2) = 0.5244 s.
Therefore, it takes approximately 0.5244 seconds for the book to reach the elevator floor.
(b) To find the book's speed relative to you when it hits the elevator floor, we can use the equation:
v = g * t,
where v is the speed, g is the acceleration due to gravity, and t is the time. Since the elevator is descending at a constant speed, the book will also have the same downward velocity as the elevator, which is 3.4 m/s.
Therefore, the book's speed relative to you when it hits the elevator floor is 3.4 m/s.
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