If home plate is 60 feet 6 inches from the pitchers mound and if the pitchers release point is 6 feet in front of the mound, how long does it take a fastball thrown at 90 miles per hour to reach home plate?
2007-03-09 12:14:04 · 7 answers · asked by gca18 1 in Sports ➔ Baseball
it should be about 1-1.5 seconds
2007-03-09 12:42:37 · answer #1 · answered by Anonymous · 0⤊ 2⤋
the 1st 2 solutions are purely incorrect. you won't have the ability to apply the appropriate attitude physique of techniques because of the fact in case you draw a line from first to one/3, the pitchers mound is in front of the line, not at as quickly as on it, and consequently you do not have a suited triangle. Mr. B has the appropriate physique of techniques, and in all probability the single your instructor is waiting so you might apply on the homework situation, different than that Mr. B did the maths incorrect. He forgot to sq. the ninety foot factor and that i'm uncertain the place the cosine went except he assumed that Cos (40 5) = a million, that's misguided, because of the fact Cos (40 5) = .707... you additionally can use suited triangles to first compute the right distance from abode to 2nd base. Then create a suited triangle between the middle factor of the infield, the pitchers mound, and popular base, and use yet another suited triangle to compute the right distance from the mound to first, yet be conscious that the mound isn't on the factor of the appropriate attitude, its on the factor on the different end of an extremely short factor and an attitude particularly below ninety stages. TomJC already did this by using the two techniques, yet the two would desire to yeild the right comparable result except you have rounding errors compounded by using your math (that's in all probability the case). I purely theory i'd chime in and clarify the comparable element in a diverse way in case it helps. yet i think of TomJC merits best answer.
2016-12-18 09:36:32 · answer #2 · answered by ? 4 · 0⤊ 0⤋
Faster than Ralph Kiner can mispronounce Galarraga.
2007-03-12 07:55:58 · answer #3 · answered by Bob Mc 6 · 0⤊ 0⤋
less than a second. But it takes to seconds for the ump to call strike 3 on you and another half a second to say your out.
2007-03-12 21:59:21 · answer #4 · answered by skisram 4 · 0⤊ 0⤋
90mi/hr x 1hr/60min x 1hr/60sec x 5280ft/1mi = 132ft/sec
54.6/132 = .417 seconds (round off to thousandths.)
simple math.
2007-03-09 13:53:11 · answer #5 · answered by what? 7 · 0⤊ 0⤋
(90 miles/hr) x (1 hr/60 min) x (1 min/60 sec) x (5280 ft/mile)
= 132 ft/sec
(54.5 ft )/ (132 ft/sec)
=0.4128 sec
It's simple math really :)
2007-03-09 12:43:13 · answer #6 · answered by mathman 2 · 4⤊ 0⤋
1.61 seconds
2007-03-09 17:56:55 · answer #7 · answered by R Dubbya 1 · 0⤊ 0⤋