Problem #2 Consider the following question about a baseball thrown by a pitcher,

Question

Problem #2 Consider the following question about a baseball thrown by a pitcher,

from the mound, and caught by the catcher, at home plate, 60 feet and 6 inches away. Assume the baseball is thrown at a speed of 101 miles per hour and that it moves in a straight line path without changing speed (we make these assumptions to keep the problem simple).

(a) If in the process of the throw, before the ball has left the pitcher’s hand, the ball moves over a distance of 1.50 m, what is the acceleration of the ball? Assume that the ball is uniformly accelerated.

(b) How long will it take the ball to arrive at the catcher’s position?

Explanation / Answer

given

horizontal distance travelled, x = 60 feet 6 inch

= 60*0.3048 + 6*2.54*10^-2

= 18.44 m

distance travelled in the hand, d = 1.5 m

initial velocity, u = 0

velocity of ball when it leaves pitcher hand, v = 101 mile/hour

= 101*1.609*5/18 m/s

= 45.14 m/s

a) let a is the acceleration of the ball.

Apply, v^2 - u^2 = 2*a*d

==> a = (v^2 - u^2)/(2*d)

= (45.14^2 - 0^2)/(2*1.5)

= 679.2 m/s^2

b) let t1 is the time taken in the hand.

t1 = (v - u)/a

= (45.14 - 0)/679.2

= 0.0664 s

time taken in the air, t = x/v

= 18.44/45.14

= 0.4085 s

so, total time taken, t = t1 + t2

= 0.0664 + 0.4085

= 0.475 s

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