# A major-league pitcher can throw a ball in excess of 39.1 m/s. If a ball is thrown horizontally at this spe?

A major-league pitcher can throw a ball in excess of 39.1 m/s. If a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?

Relevance

Dx = horizontal distance traveled = 17.0 meters

Vx = horizontal speed = 39.1 m/sec = 87.46 mi/h

Time of Flight = t = (Dx) ⁄ (Vx) = (17.0) ⁄ (39.1) = 0.43478 sec

h = height relative to initial height

Vyi = initial vertical speed = 0 ... (ball thrown horizontally)

h = (½) • g • t² + (Vyi) • t

h = (½) • (-9.81) • (0.43478)²

h = -  0.927 meters  ...  almost a meter drop-off

• Login to reply the answers
• Anonymous
4 years ago

i'm now no longer partial to T20 regardless of whether it fairly is IPL or the enormous Bash. regrettably, IPL is being pulled below without decrease than 4 game enthusiasts arrested this week, an worldwide umpire stood down.

• Login to reply the answers