An athlete puts a​ 16-lb shot at an angle of 45degrees to the horizontal from 6.5 ft above the ground at an initial speed of 38 ​ft/sec. How long after launch and how far from the athlete does the shot​ land?

We can solve this problem by analyzing the motion of the shot using kinematic equations for projectile motion. Here's a step-by-step breakdown:

### Given dataWe can solve this problem by analyzing the motion of the shot using kinematic equations for projectile motion. Here's a step-by-step breakdown:

### Given 
- Weight of the shot: 16 lb (not needed for kinematics, but for general context).
- Launch angle, \(\theta = 45^\circ\).
- Height of launch above the ground: \(h_0 = 6.5 \, \text{ft}\).
- Initial speed, \(v_0 = 38 \, \text{ft/sec}\).
- Acceleration due to gravity, \(g = 32.2 \, \text{ft/sec}^2\).

We'll calculate two things:
1. **Time of flight** (how long the shot is in the air).
2. **Horizontal distance traveled** (range) before it hits the ground.

### Step 1: Time of flight

The time of flight can be found by solving the vertical motion equation. The vertical position at any time \(t\) is given by:

\[
y(t) = h_0 + v_0 \sin(\theta) t - \frac{1}{2} g t^2
\]

We are looking for the time \(t_f\) when the shot reaches the ground, i.e., when \(y(t_f) = 0\). So, we set the equation equal to zero:

\[
0 = 6.5 + (38 \sin 45^\circ)t - \frac{1}{2}(32.2)t^2
\]

First, calculate \(v_0 \sin(45^\circ)\):

\[
v_0 \sin 45^\circ = 38 \times \frac{\sqrt{2}}{2} \approx 26.87 \, \text{ft/sec}
\]

Now the equation becomes:

\[
0 = 6.5 + 26.87t - 16.1t^2
\]

This is a quadratic equation in the form:

\[
-16.1t^2 + 26.87t + 6.5 = 0
\]

We can solve this quadratic equation using the quadratic formula:

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

where \(a = -16.1\), \(b = 26.87\), and \(c = 6.5\).

Let's calculate the time of flight.The time of flight is approximately \(1.88\) seconds.

### Step 2: Horizontal distance traveled

Now, to find how far the shot travels, we use the horizontal motion equation. The horizontal distance \(x(t)\) is given by:

\[
x(t) = v_0 \cos(\theta) t
\]

Using \(v_0 \cos(45^\circ) = 38 \times \frac{\sqrt{2}}{2} \approx 26.87 \, \text{ft/sec}\), and the time of flight \(t_f = 1.88\) seconds, we can calculate the range:

\[
\text{Range} = v_0 \cos(45^\circ) \times t_f
\] 

Let’s calculate this horizontal distance.The horizontal distance traveled by the shot is approximately \(50.6\) feet.

### Summary:
- **Time of flight**: \(1.88\) seconds.
- **Horizontal distance**: \(50.6\) feet.

Would you like further details or have any questions?

Here are five related questions for further exploration:
1. How would the time of flight change if the initial speed were increased?
2. What is the maximum height reached by the shot during its flight?
3. How would changing the launch angle affect the horizontal distance?
4. How does air resistance impact the range in real-world scenarios?
5. What happens if the shot is launched from ground level instead of 6.5 ft?

**Tip:** To maximize horizontal distance in projectile motion, the optimal launch angle in a vacuum is \(45^\circ\).

An athlete puts a​ 16-lb shot at an angle of 45 degrees to the horizontal from 6.5 ft above the ground at an initial speed of 38 ​ft/sec. How long after launch and how far from the athlete does the shot​ land?

Solve a projectile motion problem involving a 16-lb shot launched at 45 degrees from 6.5 ft above the ground with an initial speed of 38 ft/sec. Learn how to calculate the time of flight and horizontal distance traveled using kinematic equations and the quadratic formula. Suitable for students in Grades 10-12.

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