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2 years ago

What is the answer Whoever gets the answer right gets brainliest

Mathematics

1 answer:

White raven [17]2 years ago

8 0

Answer:

Step-by-step explanation:

8f=10

f=1 1/4

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A ball is projected horizontally from the top of a building. One second later, another ball is projected horizontally from the s

ehidna [41]

Answer:

t = 0 at the start of the projection

Step-by-step explanation:

To solve this we need to find the distance between the 2 positions at any given time, then solve for the least distance

Let t be the time of the 2nd ball, so t + 1 is the time of the first ball

Let g be the gravitational acceleration, v be the horizontal velocity

the y coordinates of the first and 2nd balls

$y_1 = -g(t+1)^2/2$

$y_2 = -gt^2/2$

The x coordinates of the 1st and 2nd balls:

$x_1 = v(t+1)$

$x_2 = vt$

The distance between the 2 balls is

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$

$d = \sqrt{(v(t + 1) - vt)^2 + (-g(t+1)^2/2 - (-gt^2/2))^2}$

$d = \sqrt{(vt + v - vt)^2 + g/4(t^2 - (t^2 + 2t + 1))^2}$

$d = \sqrt{v^2 + (g/4)(-2t-1)^2}$

$d = \sqrt{v^2 + (g/4)(2t+1)^2}$

As both v and g are constant and cannot be changed, d is minimum when (2t + 1) is minimum, which happens only when t is minimum = 0

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2 years ago