

 A football is punted from the ground, it passes through the apex of its flight, and it begins its downward decent, as shown. When its height above the field is h=2.6m
, the horizontal and vertical components of its velocity are observed to have the values vhorizontal=19.5m/s
 and vvertical=1.5m/s
, respectively. Use a Cartesian coordinate system with the origin located at the initial position of the football. Ignore air resistance. Calculate the time, in seconds, from when the football left the ground until it is descending at a height of h=2.6m
. 

To solve this problem, we'll break it into steps and apply basic principles of kinematics. Since air resistance is ignored, we'll treat it as a simple projectile motion problem.

### Step 1: Analyze the vertical motion
We are given:
- The height \( h = 2.6 \, \text{m} \),
- The vertical velocity at that point is \( v_{\text{vertical}} = -1.5 \, \text{m/s} \) (the negative sign is because the football is descending),
- The initial vertical velocity \( v_{0y} \) is unknown,
- The acceleration in the vertical direction is \( a_y = -9.8 \, \text{m/s}^2 \) (due to gravity).

We can use the following kinematic equation to find the time \( t \) from when the football was punted until it reaches this height during its downward descent:

\[
v_y = v_{0y} + a_y t
\]

Substitute the known values:

\[
-1.5 = v_{0y} - 9.8 t
\]

\[
v_{0y} = 9.8 t - 1.5
\tag{1}
\]

Next, we apply another kinematic equation for vertical displacement:

\[
y = v_{0y} t + \frac{1}{2} a_y t^2
\]

Substitute the known height \( h = 2.6 \, \text{m} \):

\[
2.6 = v_{0y} t - \frac{1}{2} (9.8) t^2
\]

Substitute \( v_{0y} \) from Equation (1):

\[
2.6 = (9.8 t - 1.5) t - 4.9 t^2
\]

Simplify the equation:

\[
2.6 = 9.8 t^2 - 1.5 t - 4.9 t^2
\]

\[
2.6 = 4.9 t^2 - 1.5 t
\]

Rearrange the terms:

\[
4.9 t^2 - 1.5 t - 2.6 = 0
\]

This is a quadratic equation in the form \( at^2 + bt + c = 0 \), where:
- \( a = 4.9 \),
- \( b = -1.5 \),
- \( c = -2.6 \).

### Step 2: Solve the quadratic equation
We can now solve for \( t \) using the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Substitute the values for \( a \), \( b \), and \( c \):

\[
t = \frac{-(-1.5) \pm \sqrt{(-1.5)^2 - 4(4.9)(-2.6)}}{2(4.9)}
\]

Let's compute this.The two solutions for \( t \) are:

- \( t_1 = 0.897 \, \text{seconds} \)
- \( t_2 = -0.591 \, \text{seconds} \) (This is not physically meaningful as time cannot be negative.)

Thus, the time from when the football was punted until it reaches a height of \( h = 2.6 \, \text{m} \) during its downward descent is approximately \( t = 0.897 \, \text{seconds} \).

Would you like more details on any part of this solution?

### 5 Related Questions:
1. How can we calculate the maximum height the football reaches?
2. What was the initial vertical velocity of the football?
3. How long was the football in the air before hitting the ground?
4. What was the total horizontal distance traveled by the football?
5. How would the trajectory change if air resistance were considered?

### Tip:
For projectile motion problems, always break the motion into horizontal and vertical components to simplify calculations.

A football is punted from the ground, it passes through the apex of its flight, and it begins its downward descent. When its height above the field is h=2.6m, the horizontal and vertical components of its velocity are vhorizontal=19.5m/s and vvertical=1.5m/s. Ignore air resistance. Calculate the time, in seconds, from when the football left the ground until it is descending at a height of h=2.6m.

Learn how to calculate the time it takes for a football to descend to a height of 2.6m during projectile motion with given velocity components. This problem covers kinematic equations, quadratic formulas, and is suitable for students in Grades 9-12.

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