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A cat is chasing a mouse. The mouse runs in a straight line at a speed of 1.5 m/s
.If the cat leaps off the floor at a 30 degree angle and a speed of 4.0 m/s, at what distance behind the mouse should the cat leap in order to land on the poor mouse?

2006-12-10 07:43:21 · 1 answers · asked by MattS 1 in Science & Mathematics Physics

1 answers

The cat's vertical velocity Vy = 4*sin(30). He will be airborne for t=2Vy/g. He gains on the mouse at Vxrel = 4*cos(30)-1.5 m/s, and thus gains Vxrel*t m in the jump, which is the distance his launch point should be behind the mouse.

2006-12-10 08:55:49 · answer #1 · answered by kirchwey 7 · 0 0

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