In 1947 Bob Feller, a pitcher for the Cleveland Indians, threw a baseball across the plate at 98.6 mph or 44.1 m/s. For many years this was the fastest pitch ever measured. If Bob had thrown the pitch straight up up, how high would it have gone? Please tell me how to solve this!
2007-09-29 16:06:16 · 2 answers · asked by James J 1 in Science & Mathematics ➔ Physics
Thank you guys for giving those terrific answers and showing me how you got them.
2007-09-29 16:22:47 · update #1
H=0.5 gt^2
V=V0 - gt
t=V0/g
H=0.5 g (V0/g)^2
finally
H= [(V0)^2]/(2g)
H= [44.1 ^2]/(2 x 9.81)
H=99.1 m
2007-09-29 16:13:20 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
Vf = Vi + at = 44.1 - 9.8 (t) and Vf = 0 at the top, so
0 = 44.1 - 9.8t
9.8t = 44.1
t = 44.1/9.8
and then you use
y2 - y1 = v1t + 1/2 * a * t^2 and let y1 = 0
so y2 = 44.1 * (44.1/9.8) + 1/2 * (-9.8) * (44.1/9.8)^2 = 99.2m
2007-09-29 16:12:01 · answer #2 · answered by hrothgar 6 · 0⤊ 0⤋