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By Mo222 (Mo222) on Wednesday, October 01, 2003 - 04:04 pm: Edit |

How fast must a ball be thrown upward to reach a height of 12 m?

By Euphoria (Euphoria) on Wednesday, October 01, 2003 - 04:14 pm: Edit |

15.3 m/s

By Geniusash (Geniusash) on Wednesday, October 01, 2003 - 04:17 pm: Edit |

iv=x

fv=0

h=12

use equation, fv^2=iv^2+2a(h) (fv=final velocity, iv=initial velocity, a=acceleration, h=height)

0=x^2+2(-9.8)(12)

235.2=x^2

x=sqr235.2 (no calc, sorry)

By Perry2006 (Perry2006) on Wednesday, October 01, 2003 - 04:21 pm: Edit |

Hope this helps.

Method 1 (Comprehensive)

0=vi + -9.81t

t=vi/9.81

12=vi (vi/9.81) + .5(-9.81)(vi/9.81)^2

solve for vi.

Method 2 (Easy plug-in - Recommended)

Vf^2 = Vi^2 + 2aX

0^2 = vi^2 + 2(-9.81)(12)

solve for vi.

By Mo222 (Mo222) on Wednesday, October 01, 2003 - 04:26 pm: Edit |

Thanks --

The thing I don't get is why you are using -9.81 m/s^2 for the acceleration. How can all objects that are thrown up decellerate at a constant rate? Dosent it matter how strong the push upwards is?

By Fairyofwind (Fairyofwind) on Wednesday, October 01, 2003 - 04:31 pm: Edit |

The standard way to do it is: the derivative of position is velocity. Ther derivative of velocity is acceleration. So integrate. a=-9.8, v=-9.8t+v0, s=-4.9t^2+v0t+s0.

Solve -9.8t+v0=0, t=v0/9.8, -4.9(v0/9.8)^2+(v0)(v0/9.8)+s0=12, s0=0, so v0=15.3.

15.3 m/s.

Alternatively, you can use the cheesy little kinematic equation that doesn't work when acceleration isn't constant: V_f^2=V_i^2+2a(delta S)

By Geniusash (Geniusash) on Wednesday, October 01, 2003 - 04:33 pm: Edit |

No, because the pull of gravity is the same on everything, no matter how fast it's going.

By Mo222 (Mo222) on Wednesday, October 01, 2003 - 04:38 pm: Edit |

Yea but it's going up so its acceleration will be made less by gravity, but woulden't the rate at which it goes upward still depend on how strong the push up is? If there is a really strong person who throws it then it will be affected by gravity the same as if a really weak person throws it, but the acceleration will be different. What am I thinking wrong?

By Fairyofwind (Fairyofwind) on Wednesday, October 01, 2003 - 04:41 pm: Edit |

It's not acceleration. You're thinking initial velocity. When the ball leaves the person's hand, it starts off at a nonzero velocity. There is no acceleration involved there. But remember acceleration has a direction. So say it starts off at 60 m/s UPWARDS. Gravity will accelerate it downwards at 9.8 m/s^2, so that after one second, it'll be moving at 50.2 m/s (STILL UPWARDS), etc. until it becomes negative, then the ball starts moving downwards. This is why you are looking for the time the velocity = 0.

By Geniusash (Geniusash) on Wednesday, October 01, 2003 - 04:42 pm: Edit |

gravity=the acceleration

By Geniusash (Geniusash) on Wednesday, October 01, 2003 - 04:44 pm: Edit |

Fairy, have you looked at that Probability problem that's posted on this board? Am I just getting REALLY stupid?

By Mo222 (Mo222) on Wednesday, October 01, 2003 - 04:45 pm: Edit |

OK I think I get it. Thanks fairy.

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