Question

A softball base runner is trying to steal second base which means she is trying to run from first base to second base in less time than it takes the pitched ball to reach the catcher and the catcher to throw the ball to second base. (If you're not familiar with softball, find someone who is to explain what's going on) The base-runner can get from first to second base in 2.07 s. If she leaves first base as the pitcher throws a 67 mi/hr pitch which travels the 40-ft distance to the catcher, and if the catcher takes 0.65 s to catch and rethrow the ball to second base, how fast does the catcher have to throw the ball to second base to make an out? Home plate to second base is the diagonal of a square 60 ft on a side.

Answer 1

Let,

            PA = Distance between pitcher and catcher = 40 ft. (12.192 m)

            BC = Distance between 1^st and 2^nd base = 60 ft.

            AC = Distance between pitcher and 2^nd base = 60 √2 ft. (25.86 m)

            V_b = Speed of ball thrown by pitcher to catcher = 67 miles/hr. (29.95 m/s)

            T₁ = Time taken by catcher to catch and rethrow the ball = 0.65 sec.

            T₂ = Time taken by runner to reach 2^nd base from 1^st base = 2.07 sec.

Time taken by ball to reach catcher from pitcher, T₃ = PA/speed of ball

                                                                                        = (12.192 m)/(29.95 m/s)

                                                                                         = 0.407 sec.

Total time taken by ball before it rethrown by catcher, T₄ = T₁ + T₃

= 0.65 + 0.407 =1.057 sec.

Let catcher throw the ball at a certain speed (let V₀ ) such that ball reach the 2^nd base simultaneously with runner.

Hence, Time remaining to reach the 2^nd base by ball, T₅ = T₂ – T₄

= 2.07 – 1.057 = 1.013 sec.

So, V_o = AC/T₅ = 25.86/1.013 = 25.53 m/s.

        V₀ = 57.11 mi/hr

Thus, To make an out catcher have to throw the ball at a speed greater than 57.11 mi/hr to 2^nd base.