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[请教]问一个物理的问题，关于BASEBALL的！
beidog@2006-04-09 16:31
Assuming a baseball player’s home run leaves his bat at an angle of 54 degrees to the horizontal and a speed of 93 miles per hour, would it travel far enough to splash down in
McCovey Cove outside AT&T Park？
Assume McCovey Cove is 420 feet away. Also ignore any aerodynamic effects from the spin of the ball or air resistance (so the ball’s flight will be symmetrical), and assume that the water is at the same elevation as the batter's box. Use g = 32.2 feet per second squared for the acceleration due to gravity.
calculate the number of feet traveled to one decimal place？
Also calculate the maximum height (in feet) that the ball reaches and how long (in seconds) it's in the air？
Finally, assuming that this is an average home run swing for
the player, and that his average home run ball in reality usually travels 400 feet, calculate how much percentage effect air resistance has on a home run？
That is, the home run described above would normally travel 400 feet, but you came up with a different distance because you ignored air resistance. So by what percentage does air resistance decrease the length of the home run？ (This is a percentage error calculation, where the true value is 400 feet, and the calculated value is whatever number you calculated for the distance？
问题不太明白，不太了解base ball 的说－－
首先那个home run, 然后McCovey Cove 是个目标，或者地方之类吧？
还有个为什么那里有个WATER－－
问题也有N个－－了解base ball和Physics 指教一下吧！
iou1985@2006-04-10 03:56
偶仔细想了一想，其实好像也不是那么复杂
先把93MPH换成 136.4 ft/s
然后用(1/2)m*v^2=m*g*h （这里的v是136.4*sin(45)）
得出h=144.448，h 就是 maximum height了
得到h之后，用
(1/2)*g*t^2+v*t=h 来算出t （这里的v也是136.4*sin(45)）
然后得到 t＝3， t*2=6 就是在空中的时间了
最后酸楚球走的距离，就用v*t （这里的v是 136.4*cos(45) ）
然后得到距离是 578.696
最后算空气阻力的影响，就是 (578.696-400)/578.696*100%
seiyafan@2006-04-10 06:23
homerun就是把球击出球场
Marsiss@2006-04-11 03:01
动量以及动能势能瘦恒……只记得大概，但题目不难，罗嗦了一堆
就是一球以某速度某角抛物线运动，在忽略空气阻力只考虑重力的情况，求最高点以及落地点……
jsan@2006-04-11 03:10
Ignor the water part. That is just to tell you the start and the end point is the same high.
This is a vector problem. Find the two vector, then you will know how far and how high the ball can travel.
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