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Hitting a moving target below when launched 100 m above at a 45 degree angle |
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| Dec23-12, 03:57 AM | #1 |
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Hitting a moving target below when launched 100 m above at a 45 degree angle
i'm having difficulty trying to figure out a problem i thought of.
a projectile is launched upwards at a 45 degree angle 100 m above a surface. below is a target whose horizontal distance is 5 m from the launch point. just as the projectile is launched, the target moves horizontally at a constant 9 m/s. what would the velocity of the launch have to be in order to hit the target below? so far i have come up with: 5 + 9t = vcos45t for t = time i have: (see attachment) but i'm having difficulty knowing how to incorporate it properly into the equations. i've reduced the time to: √([.02548v2 + 100]/4.9) + (.07215v) is there an easier way to go about solving this? any help would be greatly appreciated. thank you |
| Dec23-12, 07:50 AM | #2 |
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I don't think there is an easier way.
I would try to use g instead of its numerical value and the analytic value for sin(45°) as long as possible, that avoids decimal numbers. |
| Dec23-12, 01:43 PM | #3 |
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i came up with a velocity of 13.985 m/s at a time of 5.638 seconds thanks for your help...now i can sleep soundly :-) |
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