Confusing 2-D projectile motion physics problem? Please help - 10 points
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Confusing 2-D projectile motion physics problem? Please help - 10 points

[From: ] [author: ] [Date: 12-07-13] [Hit: ]
please help?-The flight time is 0.47 seconds -- about a half-second.Below the horizontal means the purse was thrown downward at an angle, not upward like most of these problems.The only thing that matters here is gravity and the initial downward velocity.......
Darren flings Lauren's purse down at an angle 23 degrees below the horizontal at a speed of 4.2 m/s, and the purse leaves his hand at a height of 2.0 above the ground. How long is the purse's total flight time?

I'm not dumb, I get how to do this, but the words BELOW THE HORIZONTAL are throwing me off... i'm not sure how the diagram would look.... or how to work with it for that matter, please help?

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The flight time is 0.47 seconds -- about a half-second. (Answer)

"Below the horizontal" means the purse was thrown downward at an angle, not upward like most of these problems. The only thing that matters here is gravity and the initial downward velocity. How far out the purse goes doesn't matter.

The initial vertical velocity is -4.2 sin 23°, with the minus sign because it was thrown downward.

So with an initial displacement of +2 meters, the formula is

0 = -4.9t^2 - 4.2 sin 23 t + 2 (the 0 is because it ends up on the ground at 0 height)

Easiest way to solve this quadratic is to multiply through by -10/7, leaving this:

0 = 7t^2 + 6 sin 23 t - 20/7

Apply the quadratic formula:

t = [-6 sin 23 ± √(36 sin^2 23 + 80)] / 14

Factor a 4 out from under the radical, and divide top & bottom by 2 to get

t = [-3 sin 23 ± √(9 sin^2 23 + 20)] / 7

You have to use the plus sign in front of the radical, and the answer is 0.47 seconds. (Answer)
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