When a soccer ball is drop-kicked into the air, the height h in feet of the ball after t seconds can be modeled by the quadratic function h(t) = -16t^2 + 32t + 4.

Question

a. When does the ball hit the ground? Round to the nearest thousandth of a second.

I used the quadratic formula for this and got the answers -.1180 seconds which doesn't make sense since you can't have negative time, and 2.118 seconds which would be the answer. The ball hit the ground after 2.118 seconds.

b. What is the highest point the ball reaches, and when?

I first used the vertex formula and got the highest point was 1 foot (which doesn't make sense to me since soccer balls go higher than 1 foot????)
and then I plugged it into the function and got 20 seconds.

So, it went to a high points of 1 foot after 20 seconds? I feel like this is wrong....

Answer 1

a. h = -16T^2 + 32T + 4 = 0.80
Divide both sides by -16:
T^2 - 2 T- 0.25 = 0,
Quadratic formula:
T = (2 +- sqrt(4 +1))/2 = 2.118 and -0.1180. Use positive value.

b. V^2 = Vo^2 + 2g*h = 0.
32^2 +(- 64)h = 0,
h = 16 Ft.
h max = 4 + 16 = 20 Ft. above gnd.

V = Vo + g*T = 0. at max. ht.
32 + (-32)T = 0,
T =

Answer 2

Correction: h = -16T^2 + 32T + 4 = 0.

Answer 3

The solutions of equation :

- 16 t² + 32 t + 4 = 0 are t1/2 = ( 2 ± √5 ) / 2

t1= ( 2 + √5 ) / 2 = 2.118034

t2 = ( 2 - √5 ) / 2 = - 0.118034

Time can't be negative so the ball will hit the ground after 2.118 sec rounded to the nearest thousandth of a second.

For vertex use the formula: t = - b / 2 a , for the t coordinate , then plug it in to find the h.

In this case: a = - 16 , b = 32 , c = 4

t = - b / 2 a = - 32 / 2 ∙ ( - 16 ) = - 32 / 32 = 1

t = 1 sec

hmax = h(1) ) = - 16 ∙ 1² + 32 ∙ 1 + 4 = - 16 + 32 + 4 = 20 ft

So, it went to a high points of 20 ft after 1 second.