Throwing a ball

Question

Suppose you throw a ball straight up from the ground with a velocity of 80 ft. per second. As the ball moves upward, gravity slows it. Eventually the ball begins to fall back to the ground. The height h of the ball after t seconds in the air is given by the quadratic function h(t) = -16t² +80t. How high does the ball go? How many seconds does it take for the ball to hit the ground?

Answer 1

Dear Haley,

 

     Let's look at the given data:  h(t)= -16 t² +80t

 

Question #1:  How high does the ball go?  (maximum height)

 

Use the vertex form for a parabola:  y= a(X-h)² +K . 

 

(If "a" is positive, the parabola opens upward, and if "a" is negative, it opens downward. Also, the vertex of the

 

parabola is (h,K).) You will want to complete the square on h(t)= -16 t² +80t .

                                                                 Completing the Square

 

1.  Factor out -16 from both terms:  -16( t² - 5t)

 

2.  Take half of the middle term constant:    -5/2   and square it:   (-5/2)²  = (25)/4

 

3.  You make a third term with it by adding it to the problem, but in a way, you also have to subtract it, so that you

 

are not breaking the math rules.

                                                                                  5/2= 2.5

Simplify

                                                                   h           K

-16( t²-5t + 25/4) +(-25/4 )(-16)=    -16 ( t- 5/2)²   + 100     Vertex Form   ( 2.5, 100)

 

Therefore, at 2.5 seconds up, the ball is at its maximum height of 100 feet.

 

Question 2:  How many seconds does it take for the ball to hit the ground?  You're looking for an intercept at the

 

ground, where the height is zero feet.

 

1.  Just factor out "-16t "  the original equation:  h(t) = -16 t² +80t =  -16t (t-5) 

 

                                         h(t)=0              -16t =0   or   t-5 =0

                                                                                                      

                                                                 t=0      or t=5 seconds   V(5,0)

 

Don't use t=0 because that was before the ball started in motion.  t=5, means that it was traveling 5 seconds when

 

it hit the ground. 

 

You may check work by using a graphing calculator; just change Y max to about 100 feet.  Use Window function.  Let X min be 0 and X max 10, y min =0 , and y max =100.  The X axis represents time (t), and the y- axis represents height of ball.  Next just change the equation to the following: y= -16X²+ 80X.  Enter it into the y function mode and press Graph.  You should be able to see the Maximum point (vertex) of a parabola opening downward.  Then press 2nd Graph, and you can see the table and check the values to see if they are the same.

 

Haley, if you like this work, please give me your vote.

Answer 2

Haley, the way to answer the question in how high the ball will go is to rewrite the quadratic equation in vertex form. This can be done by completing the square.

 

h(t)=-16t²+80t

h(t)=-16(t²-5t)

h(t)=-16(t²-5t+(5/2)²)+16*(5/2)²

h(t)=-16(t-2.5)²+16*(25/4)

h(t)=-16(t-2.5)²+100 The last term is the vertex of the inverted parabola.

Hence, the ball would go 100 feet high.

 

To answer the question in how many seconds the ball will hit the ground is to factor the equation to determine at what times to ball is on the ground.

 

h(t)=-16t(t-5) Hence, h(0)=0 and h(5)=0

 

Therefore, the ball will hit the ground in 5 seconds.