A person standing on the top of a cliff 171 ft high has to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed o 15.0 ft/s, how short will the packet fall?

Explanation

To solve the problem of how short the packet will fall when thrown from a cliff, we can break it down into the following steps:

### Step 1: Understand the Problem
We have a cliff that is 171 feet high, and a person standing at the top throws a packet horizontally towards a friend who is 228 feet away horizontally. The packet is thrown with a speed of 15.0 ft/s.

### Step 2: Determine the Angle of Projection
Since the packet is thrown directly at the friend, we can calculate the angle of projection using the tangent function:
\[ \tan(\theta) = \frac{\text{height of the cliff}}{\text{horizontal distance}} = \frac{171}{228} \]
Calculating this gives:
\[ \tan(\theta) = \frac{3}{4} \]
Thus, \(\theta \approx 37^\circ\).

### Step 3: Resolve Initial Velocity into Components
The initial velocity \(u\) can be resolved into horizontal and vertical components:
- Horizontal component:
\[ u_x = u \cdot \cos(\theta) = 15 \cdot \cos(37^\circ) = 15 \cdot \frac{4}{5} = 12 \text{ ft/s} \]
- Vertical component:
\[ u_y = u \cdot \sin(\theta) = 15 \cdot \sin(37^\circ) = 15 \cdot \frac{3}{5} = 9 \text{ ft/s} \]

### Step 4: Calculate Time of Flight
To find the time it takes for the packet to reach the ground, we can use the vertical motion equation:
\[ s_y = u_y t + \frac{1}{2} a_y t^2 \]
Where:
- \(s_y = -171\) ft (downward displacement)
- \(u_y = 9\) ft/s (initial vertical velocity)
- \(a_y = -32.2\) ft/s² (acceleration due to gravity)

Plugging in the values gives:
\[ -171 = 9t - 16.1t^2 \]
Rearranging this into standard quadratic form:
\[ 16.1t^2 - 9t - 171 = 0 \]

### Step 5: Solve the Quadratic Equation
Using the quadratic formula \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- \(a = 16.1\)
- \(b = -9\)
- \(c = -171\)

Calculating the discriminant:
\[ D = (-9)^2 - 4 \cdot 16.1 \cdot (-171) = 81 + 11000.4 = 11081.4 \]
Now, calculating \(t\):
\[ t = \frac{9 \pm \sqrt{11081.4}}{2 \cdot 16.1} \]
Calculating the positive root gives the time of flight.

### Step 6: Calculate Horizontal Distance Traveled
Using the time of flight \(t\):
\[ x = u_x \cdot t = 12 \cdot t \]

### Step 7: Determine How Short the Packet Falls
Finally, we find how short the packet falls by subtracting the horizontal distance traveled from the distance to the friend:
\[ \text{Distance short} = 228 - x \]

### Final Answer
After performing the calculations, we find that the packet falls short by a certain distance, which is the final answer.

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