A ball is thrown into the air Its height in feet t seconds l

A ball is thrown into the air. Its height (in feet) t seconds later is given by

h(t)= 128t-16t²

Based on the two equivalent forms of the function

h(t)= 128t - 16t² = t*(128-16t), answer the following questions:

a) What is the height of the ball 1 second after it has been thrown?

b) After how many seconds does the ball hit the ground?

c) At what time(s) is the ball 40 feet above the ground? If there is more than one answer, give your answer as a comma separated list of values. Note that you do not have to include units in this answer.

After______________ seconds

Find a quadratic equation for a function y=f(x) which crosses the x-axis at x= -1 and x= 6, and which crosses the y-axis at y= -12.

f(x)=________________

Find the zeros of y= 20x^2-49x+30. If there is more than one zero or a zero is repeated, enter your answers as a comma separated list.

The zeros are x=________________

Solution

1) h(t)= 128t - 16t^2 = t*(128-16t)

i) height of the ball 1 second : h(1) = 128*1 -16*1 = 112 feet

ii) ball hit the ground : h(t) =0. So, t*(128-16t) =0

so we have t=0 , t= 128/6 .Now intially ball started from ground so, t=0

and after t= 21.33 sec ball returned back to ground.

iii) ball 40 feet above the ground:

h(t) =40 ; 128t - 16t^2 = 40

Now solve the quadratic equation :  128t - 16t^2 = 40 ----> -2t^2 +16t - 5 =0 ( after dividng equation by 8)

-2t^2 +16t - 5 =0 . Now solve the quadratic equation : t = ( -16 + /- sqrt( 256 -40) )/-4

= ( 4 +/3.67) = 7.67 , 0.32 seconds

t= 0.32 seconds , 7.67 seconds

2) a quadratic equation for a function y=f(x) which crosses the x-axis at x= -1 and x= 6, and which crosses the y-axis at y= -12.

Let the quadratic equation : y= a(x- b)(x-c)

now b = 1 ; c= 6 >Wefind a from point ( 0, -12)

So, y = a( x+1)(x-6)

-12 = a( 1)(-6)

a = -2

y = -2( x+1)( x-6) = -2( x^2 -5x -6) = -2x^2 +10x +12 ( Quadratic equation)