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.
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....
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.
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 =
Correction: h = -16T^2 + 32T + 4 = 0.
To solve part a, you correctly applied the quadratic formula to find the solutions of the equation h(t) = 0:
-16t^2 + 32t + 4 = 0
However, it seems that there was a mistake in your calculations. Let's go through it step by step to find the correct answer:
Using the quadratic formula, we have:
t = (-b ± √(b^2 - 4ac)) / (2a)
For our equation -16t^2 + 32t + 4 = 0, we can identify a = -16, b = 32, and c = 4. Substituting these values into the quadratic formula, we get:
t = (-32 ± √(32^2 - 4(-16)(4))) / (2(-16))
t = (-32 ± √(1024 + 256)) / -32
t = (-32 ± √1280) / -32
Simplifying further:
t = (-32 ± √(1280)) / -32
t = (-32 ± 35.777) / -32
Now, we have two solutions:
t = (-32 + 35.777) / -32 = 3.777 / -32 ≈ -0.118 (rounded to the nearest thousandth)
t = (-32 - 35.777) / -32 = -67.777 / -32 ≈ 2.118 (rounded to the nearest thousandth)
As you mentioned, negative time doesn't make sense in this context, so we discard the negative solution. Therefore, the ball hits the ground after approximately 2.118 seconds.
Moving on to part b, to find the highest point the ball reaches, we need to find the vertex of the quadratic function. The vertex of a quadratic function in the form h(t) = at^2 + bt + c can be found using the formula:
t = -b / (2a)
In our case, a = -16 and b = 32. Substituting these values, we have:
t = -32 / (2(-16))
t = -32 / -32
t = 1
So, the highest point of the ball's trajectory occurs at t = 1 second.
Now, to determine the height at t = 1 second, we substitute t = 1 into the quadratic function:
h(1) = -16(1)^2 + 32(1) + 4
h(1) = -16 + 32 + 4
h(1) = 20
Therefore, the highest point the ball reaches is at 20 feet after 1 second, not 1 foot after 20 seconds.
To summarize:
a. The ball hits the ground approximately after 2.118 seconds.
b. The highest point that the ball reaches is 20 feet after 1 second.