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?

Answer 1

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

Answer 2

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.

Answer 3

thanx both of yall XD