Asked by Abdullah
Initially, a baseball was held by one pitcher.
The fastest measured pitched baseball left the pitcher’s hand at a speed of 45m/s. If the pitcher was in contact with the ball over a
distance of 1.5m and produced constant acceleration,
a)What acceleration did he give the ball?
b)How much time did it take him to pitch it?
The fastest measured pitched baseball left the pitcher’s hand at a speed of 45m/s. If the pitcher was in contact with the ball over a
distance of 1.5m and produced constant acceleration,
a)What acceleration did he give the ball?
b)How much time did it take him to pitch it?
Answers
Answered by
oobleck
Solve these:
v = at
s = 1/2 at^2
v = at
s = 1/2 at^2
Answered by
Damon
speed 0 to 45 m/s over 1.5 meters at constant a
how long was it in his hand?
v = a t, so v linear in time
so average speed = 45/2 =22.5 m/s
so time in hand= 1.5 /22.5 = 0.0667 seconds (ans part b)
v = a t still
45 = a (0.0667)
a = 675 m /s^2
by the way, 45m/s is 100 miles per hour. Really?
how long was it in his hand?
v = a t, so v linear in time
so average speed = 45/2 =22.5 m/s
so time in hand= 1.5 /22.5 = 0.0667 seconds (ans part b)
v = a t still
45 = a (0.0667)
a = 675 m /s^2
by the way, 45m/s is 100 miles per hour. Really?
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