Question

How far in ft must a catcher throw the ball to throw out a runner stealing second base

Answer 1

To calculate the distance beween the home plate and the second base,we will use the pythagoras theorem

Using the Pythagorean theorem,

$\text{hypotenus}^2=\text{opposite}^2+\text{adjacent}^2$

Where,

$\begin{gathered} \text{hypotenus}=x \\ \text{opposite}=90ft \\ \text{adjacent}=90ft \end{gathered}$

Substituting the values,we will have

$\begin{gathered} x^2=90^2+90^2^{} \\ x^2=8100+8100 \\ x^2=16200 \\ x=\sqrt[]{16200} \\ x=127.28ft \end{gathered}$

Convert 60ft 6in to ft we will have,

$\begin{gathered} 1in=0.0833ft \\ 6in=0.0833\times6 \\ 6in=0.5ft \\ \text{therefore, } \\ 60ft\text{ 6in}\Rightarrow(60+0.5)ft=60.5ft \end{gathered}$

Therefore, The distance of the catcher to the second base will be

$\begin{gathered} \text{distance =}127.28ft-60.5ft \\ \text{distance}=66.78ft \\ to\text{ the nearest foot} \\ \text{distance}=67ft \end{gathered}$

Hence,

The final answer is = 67ft