20 meters per second
0.95 - 1.05 secs
12.5 m/s
300cm. Ignoring the first drop, the first bounce is 50+50cm (up then down!), the second is 25+25, the third 12.5+12.5 ... and so on. If you know or accept that 1+1/2+1/4+1/8+1/16...=2 then the bounces add up to 100+50+25+12.5...=200, and then add the initial 100cm drop to make 300cm.
weight does not matter. if it has air resistance which makes a paper air plane fly straight and faster which is the point at the end than it will go faster but if its flat like a piece of paper than it will float. im just estimating but i would say it would go 500 or 600 feet maybe in between that.
Energy is being lost at several points. There is air resistance, of course, but the largest loss of energy would be due to the fact that the ball is not perfectly elastic.
You drop it once, if it comes to rest on the path, you drop it again, on the second drop if it comes to rest on the path you place it where the ball touched the grass first on the second drop.
20 meters per second
72 meters
72 meter
yes, you go to the nearset point of relief then two clucb lengths no closer to the hole then drop the ball
As long as the tennis ball is not thrust downward, yes, the tennis ball will bounce back to the same proportion of its original height, no matter how far it's dropped, as long as the height is small enough that air resistance can be ignored. The ball will eventually come to rest due to this air resistance.
1.39 Ns up
Once you hit the putt and the ball is at rest, you are allowed to walk to the hole and wait for only ten seconds. If by 10 the ball doesn't drop you must tap it in.
Answer: 66 Meters. Just had that same problem on a math mates worksheet.
No. What counts in this case is the vertical component of the velocity, and the initial vertical velocity is zero, one way or another.
Yes, you must drop the ball within the two club lengths and the ball must come to rest within the two club lengths.
0.95 - 1.05 secs