Someone drops a ball from the roof of your building.The ball drop past your window. It takes time T to pass the window's height. What is the initial speed of the ball when it first came in view?
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V1 = speed at top of window
V2 = speed at bottom of window
h = height of window
T = time of passage
[(V1 + V2) / 2] * T = h
V2 = V1 + g T
[(V1 + V1 + g T) / 2] * T = h
V1 T + (g T^2 / 2) = h
V1 = (h / T) - (g T / 2)
To determine the initial speed of the ball when it first came into view, we can use the equations of motion. From the given information, we know that the time it takes for the ball to pass the window's height is denoted as T.
The key equation we can use is the equation of motion for vertical motion:
d = vit + (1/2)at^2
Where:
- d is the distance traveled
- vi is the initial velocity
- t is the time
- a is the acceleration
In this scenario, we have two segments of the ball's motion: the portion before it reaches the window and the portion after it passes the window. Let's break it down:
1. Segment before reaching the window:
We want to find the initial velocity (vi) of the ball. Since it starts from rest, we can assume the initial velocity is zero. So the equation becomes:
d1 = (1/2)at^2
2. Segment after passing the window:
Now, we want to determine how far the ball traveled from when it was first seen until it passed the window. This distance is equal to the height of the window.
d2 = (1/2)at^2
Since the total distance traveled is the sum of the distances in both segments, we have:
d1 + d2 = h
Now we have the equation:
(1/2)at^2 + (1/2)at^2 = h
Since the acceleration due to gravity (a) is constant, we can simplify the equation to:
at^2 = 2h
Solving for the initial velocity (vi), we rearrange the equation:
vi = (2h / t^2)
Therefore, the initial speed of the ball when it first came into view is given by (2h / t^2).