Question

An arrow is shot from a height of 1.4 m toward a cliff at an angle of 60° and a velocity of 26 m/s, and determine the height of the cliff and the maximum height reached by the arrow.

Answer 1

To determine the height of the cliff and the maximum height reached by the arrow, we can use the principles of kinematics.

### Step 1: Break Down the Initial Velocity

Given:
- Initial velocity \( v_0 = 26 \, \text{m/s} \)
- Angle \( \theta = 60^\circ \)

We can break the initial velocity into its horizontal and vertical components.

**Horizontal component**:
\[
v_{0x} = v_0 \cos(\theta) = 26 \cos(60^\circ) = 26 \times 0.5 = 13 \, \text{m/s}
\]

**Vertical component**:
\[
v_{0y} = v_0 \sin(\theta) = 26 \sin(60^\circ) = 26 \times \frac{\sqrt{3}}{2} \approx 22.5 \, \text{m/s}
\]

### Step 2: Determine the Maximum Height

To find the maximum height of the arrow above the launch height, we use the formula for maximum height in projectile motion:
\[
H = \frac{v_{0y}^2}{2g}
\]
where \( g \approx 9.81 \, \text{m/s}^2 \).

Calculating \( H \):
\[
H = \frac{(22.5)^2}{2 \times 9.81} = \frac{506.25}{19.62} \approx 25.8 \, \text{m}
\]

### Step 3: Calculate the Total Maximum Height

The total maximum height (above the ground) is:
\[
\text{Total height} = \text{Launch height} + H = 1.4 \, \text{m} + 25.8 \, \text{m} \approx 27.2 \, \text{m}
\]

### Step 4: Determine the Height of the Cliff

Since the problem does not specify the distance to the cliff, we will assume that the arrow will still be traveling upward when it reaches the maximum height, and then will fall down. To know if it will reach the cliff or how long it takes to reach the horizontal distance 'd' towards the cliff, we can find the time to reach maximum height.

The time to reach the maximum height is given by:
\[
t_{up} = \frac{v_{0y}}{g} = \frac{22.5}{9.81} \approx 2.29 \, \text{s}
\]

The total time the arrow is in the air (as it goes up and then comes back down to height 1.4 m) is twice the time taken to reach peak height:
\[
t_{total} = 2 \cdot t_{up} = 2 \cdot 2.29 \approx 4.58 \, \text{s}
\]

### Horizontal Distance

The horizontal distance traveled during this time is:
\[
d = v_{0x} \cdot t_{total} = 13 \, \text{m/s} \times 4.58 \, \text{s} \approx 59.5 \, \text{m}
\]

### Final Calculations and Heights

Overall,

- The **maximum height reached by the arrow** above the ground is approximately **27.2 meters**.
- The height of the cliff is not directly calculated from the given information as it requires the distance (horizontal range) to the cliff. If the cliff is at that distance, it would be the maximum height minus the launch height (1.4 m).

If we denote the cliff height as \( h_c \):
\[
h_c = \text{Total Height} - 1.4 = 27.2 - 1.4 \approx 25.8 \, \text{m}
\]

So if the cliff is \(59.5 \text{m}\) away in horizontal distance, it’s height would need to be evaluated based on projectile equations or additional information.

In general terms:
- **Maximum height above ground:** 27.2 m
- **Cliff height (assumed to meet the upward motion and drop):** 25.8 m above the launch point.

If the cliff is directly where it drops (vertical line), further calculations based on trajectory would be necessary to specify.