A ball is thrown upward from the ground with an initial speed of 26.7 m/s; at the same instant, a ball is dropped from a building 13.1 m high. After how long will the balls be at the same height?
Solve this equation for t:
(Thrown) Ball 1 height = (Dropped) Ball 2 height
26.7 t - 4.9 t^2 = 13.1 - 4.9 t^2
The acceleration terms, -4.9 t^2, cancel out
t = 13.1/26.7 = ___ seconds
Interesting result, and one that hints at Einstein's principle of equivalence.
In a free-falling coordinate system the dropped ball doesn't move, and the thrown ball reaches it at constant speed. Gravity seems to disappear
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To find the time it takes for both balls to be at the same height, we need to set up and solve two separate equations for each ball.
For the ball thrown upward:
1. The initial velocity (u) is +26.7 m/s (upward).
2. The acceleration due to gravity (a) is -9.8 m/s² (downward).
3. The initial height (h) is 0 m (ground level).
Using the equation h = ut + (1/2)at² to find the time, we have:
0 = (26.7)t + (1/2)(-9.8)t²
For the ball dropped from the building:
1. The initial velocity (u) is 0 m/s (ball dropped from rest).
2. The acceleration due to gravity (a) is -9.8 m/s² (downward).
3. The initial height (h) is 13.1 m (building height).
Using the equation h = ut + (1/2)at² to find the time, we have:
13.1 = 0t + (1/2)(-9.8)t²
We now have two equations:
0 = 26.7t + (-4.9)t² (equation 1)
13.1 = -4.9t² (equation 2)
Simplifying equation 2:
4.9t² = -13.1
t² = -13.1 / 4.9
t² ≈ -2.6735
Since time cannot be negative, we can disregard the negative solution.
Taking the square root of both sides gives us:
t ≈ √(-2.6735)
t ≈ 1.6346
The time it takes for the balls to be at the same height is approximately 1.6346 seconds.