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A baseball is hit with an initial upward velocity of 70 feet per second from a height of 4 feet above the ground. The equation h=−16t2+70t+4 models the height in feet t seconds after it is hit. After the ball gets to its maximum height, it comes down and is caught by another player at a height of 6 feet above the ground. About how long after it was hit does it get caught?

Bunnie  Nov 28, 2017

Best Answer 

 #1
avatar+5589 
+3

h  =  -16t2 + 70t + 4

 

We want to know how many seconds it takes for the ball to be 6 feet above the ground.

We want to know the value of  t  when  h = 6 .  So plug in  6  for  h  and solve for  t .

 

6  =  -16t2 + 70t + 4       Subtract  6  from both sides of the equation.

0  =  -16t2 + 70t - 2        Let's solve this quadratic equation using the quadratic formula.

 

t  =  \(\frac{-70\pm\sqrt{70^2-4(-16)(-2)}}{2(-16)}\)

 

t  =  \(\frac{-70\pm\sqrt{4900-128}}{-32}\)

 

t  =  \(\frac{-70\pm\sqrt{4772}}{-32}\)

 

t  =  \(\frac{-70+\sqrt{4772}}{-32}\)  ≈  0.029                    or                    t  =  \(\frac{-70-\sqrt{4772}}{-32}\)  ≈  4.346

 

So we know that

at 0.029 seconds, the ball is 6 feet above the ground and on its way up,  and

at 4.346 seconds, the ball is 6 feet above the ground and on its way down.

 

The player will catch the ball on its way down, at about 4.346 seconds.   smiley

hectictar  Nov 28, 2017
Sort: 

1+0 Answers

 #1
avatar+5589 
+3
Best Answer

h  =  -16t2 + 70t + 4

 

We want to know how many seconds it takes for the ball to be 6 feet above the ground.

We want to know the value of  t  when  h = 6 .  So plug in  6  for  h  and solve for  t .

 

6  =  -16t2 + 70t + 4       Subtract  6  from both sides of the equation.

0  =  -16t2 + 70t - 2        Let's solve this quadratic equation using the quadratic formula.

 

t  =  \(\frac{-70\pm\sqrt{70^2-4(-16)(-2)}}{2(-16)}\)

 

t  =  \(\frac{-70\pm\sqrt{4900-128}}{-32}\)

 

t  =  \(\frac{-70\pm\sqrt{4772}}{-32}\)

 

t  =  \(\frac{-70+\sqrt{4772}}{-32}\)  ≈  0.029                    or                    t  =  \(\frac{-70-\sqrt{4772}}{-32}\)  ≈  4.346

 

So we know that

at 0.029 seconds, the ball is 6 feet above the ground and on its way up,  and

at 4.346 seconds, the ball is 6 feet above the ground and on its way down.

 

The player will catch the ball on its way down, at about 4.346 seconds.   smiley

hectictar  Nov 28, 2017

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