In college softball, the distance from the pitcher's mound to the batter is 43 feet. If the ball leaves the bat at 110 mph , how much time elapses between the hit and the ball reaching the pitcher?

Respuesta :

Explanation:

It is given that,

The distance from the pitcher's mound to the batter is 43 feet, d = 43 feet = 0.00814 miles

Speed with which ball leaves the ball, v = 110 mph

Let t is the time elapses between the hit and the ball reaching the pitcher. It is given by :

[tex]t=\dfrac{d}{v}[/tex]

[tex]t=\dfrac{0.00814}{110}[/tex]

t = 0.000074 hours

or

t = 0.2664 seconds

So, the time between the hit and the ball reaching the pitcher is 0.2664 seconds. Hence, this is the required solution.

The time taken should be 0.000074 hours or  0.2664 seconds.

Calculation of the time taken:

Here we assume the time be t

And, The distance from the pitcher's mound to the batter is 43 feet, d = 43 feet = 0.00814 miles

So, the following formula should be used.

[tex]= 0.00814 \div 110[/tex]

= 0.000074 hours or  0.2664 seconds.

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