Question

One of the fastest recorded pitches major-league baseball, thrown by nolan ryan in 1974, was clocked at 100.8 mi/hr. If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically by the time it reached home plate 60.0 ft away?

Answer 1

Answer:

ball will fall down by y = 2.63 ft

Explanation:

As we know that fastest pitch for major league is given as

$v = 100.8 mph$

here we know that

$1 mph = 1.467 ft/s$

now we have

$v = 100.8 mph = 100.8 \times 1.467 ft/s$

$v = 147.87 ft/s$

now the distance of home plate is given as

$d = 60 ft$

so here the time taken by the ball top reach home plate is given as

$t = \frac{d}{v}$

$t = \frac{60}{147.87}$

$t = 0.406 s$

Now the displacement of ball in vertical direction is given as

$y = \frac{1}{2}gt^2$

$y = \frac{1}{2}(32)(0.406)$

$y = 2.63 ft$

Answer 2

Answer:

 The ball fall vertically 2.69 ft by the time it reached home plate 60.0 ft away.

Explanation:

    Fastest recorded pitches major-league baseball, thrown by nolan ryan in 1974 = 100.8 mi/hr = 44.8 m/s

    The horizontal distance to home plate = 60.0 ft = 18.288 m

  We have the horizontal velocity = 44.8 m/s

    So time taken = 18.288/44.8 = 0.408 seconds.

  The distance traveled by baseball vertically is found out by equation $s=ut+\frac{1}{2} at^2$

  Here u =0m/s, a = 9.81 $m/s^2$ and t = 0.408 s

 Substituting

     $S = 0*0.408+\frac{1}{2} *9.81*0.408^2 = 0.82 m\\ \\S = 0.82 m= 2.69 ft$

So vertical distance traveled = 0.82 m = 2.69 ft