Ice-cream Posted March 14, 2005 Posted March 14, 2005 I need some explanantion on this question. A crow drops a shell from a height half the height from which a seagull drops a shell. When the crow's shell hits the rocks below, is it moving half as fast, less than half as fast or more than half as fast the seagull's shell? The answer is: more than half as fast Can any1 explain why? Also, I need help with this question. Bills steps off a 3m high diving board and drops to the water below. At the same time Ted jumps upward with the speed of 4.2 m/s from a 1m high diving board. choosing the origin to be at the water's surface, and upward to be the positive direction, write x-versus-t equations for the motion for both bill and ted. Repeat this question, but this time using the origin at 3m above the water and with downward as the positive direction. I'm really having trouble writing the equations becuase i don't think the heights affect anything....so im not sure...can any1 give me some help? thanx
Sayonara Posted March 14, 2005 Posted March 14, 2005 Descriptive thread title = higher probability of getting help.
The Rebel Posted March 14, 2005 Posted March 14, 2005 v^2 is porportional to height, so v is proportional to sqrt of h So a drop that is 1/2 the other, it will finish with a speed of sqrt 1/2 the other. For the second question: Use s=ut + 0.5at^2 for a varying t, remembering that Ted's acceleration will change direction when 0 = u^2 + 2as
Callipygous Posted March 14, 2005 Posted March 14, 2005 v^2 is porportional to height' date=' so v is proportional to sqrt of h So a drop that is 1/2 the other, it will finish with a speed of sqrt 1/2 the other.[/quote'] put another way, speed increases proportional to time. while they both reach the same speed in the time it takes the lower one to hit. during the next half of the fall for the higher one, its already moving pretty fast, so it covers the rest of the distance faster. it doesnt have twice as much time to accelerate, so it doesnt reach twice the speed.
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