the 1st term should be -16t^2 , it reflects gravity's action during freefall
40 is the initial upward velocity of the ball
plug in 22 for h , and solve for t
The motion of a ball scooped by a field hockey player can be modeled by h= -16t + 40t, where t is the time in seconds and h is the height of the ball. Will the ball ever reach 22 feet?
I don't understand how to solve this. Help?
6 answers
So
22= -16t^2 + 40t
Subtract 22
0= -16t^2 + 40t -22
Now what?
22= -16t^2 + 40t
Subtract 22
0= -16t^2 + 40t -22
Now what?
you've heard of the quadratic formula?
h = -16 t^2 + 40 t
parabola, where is vertex?
16 t^2 -40 t = -h
t^2 - 2.5 t = -h/16
t^2 - 2.5 t + 1.25^2 = -h/16 + 1.5625
(t - 1.25)^2 = -(1/16) (h-25)
vertex (top) at t = 1.25 seconds and h = 25 feet
parabola, where is vertex?
16 t^2 -40 t = -h
t^2 - 2.5 t = -h/16
t^2 - 2.5 t + 1.25^2 = -h/16 + 1.5625
(t - 1.25)^2 = -(1/16) (h-25)
vertex (top) at t = 1.25 seconds and h = 25 feet
so it passes 22 on the way up and on the way down :)
Okay thank you
6 answers