+0  
 
+1
605
1
avatar+28 

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+7266 
+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
 #1
avatar+7266 
+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

17 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.