Posted by **Selena** on Monday, June 2, 2014 at 3:54pm.

A ball is thrown into the air. The height in feet of the ball can be modeled by the equation h = -16t2 + 20t + 6 where t is the time in seconds, the ball is in the air. When will the ball hit the ground? How high will the ball go?

- Algebra -
**Damon**, Monday, June 2, 2014 at 5:28pm
16 t^2 -20 t - 6 = - h in obsolete English units

when does h = 0?

t = [ 20 +/- sqrt (400 + 384) ] /32

t = [ 20 +/- 28 ] /32

t = 48/32 = 1.5 seconds (ignore the negative t)

for vertex complete the square assuming no calculus may be used

16 t^2 -20 t = - h + 6

t^2 - (5/4) t = -h/16 + 3/16

t^2 - (5/4)t + 25/64 = -4h/64 + 37/64

(t-5/8)^2 = -(4/64) (h - 148 )

max height = 148 at t = 5/8 sec

- Algebra -
**MathMate**, Monday, June 2, 2014 at 5:44pm
Find the vertex of the parabola by completing the square:

h(t)=-16t²+20t+6

=-16(t²-2(5/8)t+25/64)+6+25/4

which means that the vertex is at

t=5/8

and it will hit the ground at h(t)=0, or

t=3/2 s (rejecting negative root).

Maximum is at the vertex, or

h(5/8)=49/4 ft.

- Algebra -
**Selena**, Monday, June 2, 2014 at 5:44pm
thanx both of yall XD

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