Question
A ball is kicked into the air from the ground with an initial velocity of 80 feet per second. When does the ball exceed 64 feet above the ground? Round to the nearest hundredth second, if necessary.(1 point) Responses Between 1 second and 4 seconds after the ball is kicked Between 1 second and 4 seconds after the ball is kicked At exactly 1 second and 4 seconds after the ball is kicked At exactly 1 second and 4 seconds after the ball is kicked More than 4 seconds after the ball is kicked More than 4 seconds after the ball is kicked Less than 1 second after the ball is kicked
Answers
GPT-4o mini
To determine when the ball exceeds 64 feet above the ground, we can use the physics formula for the height of an object in projectile motion:
\[
h(t) = -16t^2 + v_0 t + h_0
\]
where:
- \( h(t) \) is the height in feet at time \( t \) in seconds,
- \( v_0 \) is the initial velocity (80 ft/s in this case),
- \( h_0 \) is the initial height (0 feet since it is kicked from the ground),
- and the \(-16t^2\) term accounts for the acceleration due to gravity (in feet per second squared).
Substituting in the values:
\[
h(t) = -16t^2 + 80t
\]
We want to find when the height \( h(t) \) exceeds 64 feet. Thus, we set up the inequality:
\[
-16t^2 + 80t > 64
\]
First, we rearrange this to form a standard quadratic inequality:
\[
-16t^2 + 80t - 64 > 0
\]
Next, we solve the equality:
\[
-16t^2 + 80t - 64 = 0
\]
We can simplify this by dividing everything by \(-16\):
\[
t^2 - 5t + 4 = 0
\]
Now, we apply the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = 1 \)
- \( b = -5 \)
- \( c = 4 \)
Calculate the discriminant:
\[
b^2 - 4ac = (-5)^2 - 4(1)(4) = 25 - 16 = 9
\]
Now use the quadratic formula:
\[
t = \frac{-(-5) \pm \sqrt{9}}{2(1)} = \frac{5 \pm 3}{2}
\]
This gives us two solutions:
\[
t = \frac{8}{2} = 4 \quad \text{and} \quad t = \frac{2}{2} = 1
\]
Thus, the critical points are \( t = 1 \) second and \( t = 4 \) seconds.
To find when \( h(t) \) is greater than 64, we need to test the intervals defined by these critical points: \( (-\infty, 1) \), \( (1, 4) \), and \( (4, \infty) \).
1. For a test point in \( (-\infty, 1) \) (like \( t = 0 \)):
\[
h(0) = -16(0)^2 + 80(0) = 0 \quad (\text{not greater than 64})
\]
2. For a test point in \( (1, 4) \) (like \( t = 2 \)):
\[
h(2) = -16(2)^2 + 80(2) = -16(4) + 160 = -64 + 160 = 96 \quad (\text{greater than 64})
\]
3. For a test point in \( (4, \infty) \) (like \( t = 5 \)):
\[
h(5) = -16(5)^2 + 80(5) = -16(25) + 400 = -400 + 400 = 0 \quad (\text{not greater than 64})
\]
From this analysis, we conclude that the ball exceeds 64 feet between the times \( t = 1 \) second and \( t = 4 \) seconds.
Thus, the answer is:
**Between 1 second and 4 seconds after the ball is kicked.**
\[
h(t) = -16t^2 + v_0 t + h_0
\]
where:
- \( h(t) \) is the height in feet at time \( t \) in seconds,
- \( v_0 \) is the initial velocity (80 ft/s in this case),
- \( h_0 \) is the initial height (0 feet since it is kicked from the ground),
- and the \(-16t^2\) term accounts for the acceleration due to gravity (in feet per second squared).
Substituting in the values:
\[
h(t) = -16t^2 + 80t
\]
We want to find when the height \( h(t) \) exceeds 64 feet. Thus, we set up the inequality:
\[
-16t^2 + 80t > 64
\]
First, we rearrange this to form a standard quadratic inequality:
\[
-16t^2 + 80t - 64 > 0
\]
Next, we solve the equality:
\[
-16t^2 + 80t - 64 = 0
\]
We can simplify this by dividing everything by \(-16\):
\[
t^2 - 5t + 4 = 0
\]
Now, we apply the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- \( a = 1 \)
- \( b = -5 \)
- \( c = 4 \)
Calculate the discriminant:
\[
b^2 - 4ac = (-5)^2 - 4(1)(4) = 25 - 16 = 9
\]
Now use the quadratic formula:
\[
t = \frac{-(-5) \pm \sqrt{9}}{2(1)} = \frac{5 \pm 3}{2}
\]
This gives us two solutions:
\[
t = \frac{8}{2} = 4 \quad \text{and} \quad t = \frac{2}{2} = 1
\]
Thus, the critical points are \( t = 1 \) second and \( t = 4 \) seconds.
To find when \( h(t) \) is greater than 64, we need to test the intervals defined by these critical points: \( (-\infty, 1) \), \( (1, 4) \), and \( (4, \infty) \).
1. For a test point in \( (-\infty, 1) \) (like \( t = 0 \)):
\[
h(0) = -16(0)^2 + 80(0) = 0 \quad (\text{not greater than 64})
\]
2. For a test point in \( (1, 4) \) (like \( t = 2 \)):
\[
h(2) = -16(2)^2 + 80(2) = -16(4) + 160 = -64 + 160 = 96 \quad (\text{greater than 64})
\]
3. For a test point in \( (4, \infty) \) (like \( t = 5 \)):
\[
h(5) = -16(5)^2 + 80(5) = -16(25) + 400 = -400 + 400 = 0 \quad (\text{not greater than 64})
\]
From this analysis, we conclude that the ball exceeds 64 feet between the times \( t = 1 \) second and \( t = 4 \) seconds.
Thus, the answer is:
**Between 1 second and 4 seconds after the ball is kicked.**