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Sam tosses a ball horizontally off a footbridge at 3.1 m/s. How much time passes after he releases it until its speed doubles?
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Sam tosses a ball horizontally off a footbridge at 4.8 m/s. How much time passes after he releases it until its speed doubles?
Top answer:
When the speed has doubled, the vertical component of speed will be sqrt[(9.6)^2 - (4.8)^2] = 8.31
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Sam tosses a ball horizontally off a footbridge at 3.2 m/s. How much time passes after he releases it until its speed doubles?
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The vertical velocity component Vy would have to increase to sqrt3 times the initial horizontal
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i posted this question before however i did not understand drwls explanation. anybody else want to try ?
Sam tosses a ball
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Well, it is easy if one considers energy. if velocity doubles, then velocity quadruples Vf^2=4Vi^2
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umm i need help...which equation would i use to solve the probnlem and how do i go about finding the answer?
Here it is: Sam
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I answered this earlier today. http://www.jiskha.com/display.cgi?id=1292877672
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Sam tosses a ball horizontally off a footbridge at 4.3 m/s. How much time passes after he releases it until its speed doubles?
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