To find the height of the cliff, we can focus on the vertical motion of the projectile. We know the following information:

• Initial vertical velocity: $v_{0y} = 175 \, \text{m/s}$
• Time of flight: $t = 40.0 \, \text{s}$
• Gravitational acceleration: $g = 9.8 \, \text{m/s}^2$
• The projectile ends up at a point lower than its starting height (since it falls into the sea), which means the vertical displacement will give us the height of the cliff.

Step 1: Equation for vertical displacement

The vertical displacement $y$ after time $t$ can be calculated using the kinematic equation:

$y = v_{0y} \cdot t - \frac{1}{2} g \cdot t^2$

where:

• $v_{0y} = 175 \, \text{m/s}$
• $t = 40.0 \, \text{s}$
• $g = 9.8 \, \text{m/s}^2$

Step 2: Calculate the vertical displacement

Now, let's plug in the values and calculate the displacement.

$y = (175 \, \text{m/s}) \cdot (40.0 \, \text{s}) - \frac{1}{2} (9.8 \, \text{m/s}^2) \cdot (40.0 \, \text{s})^2$

Let's calculate this:The vertical displacement $y$ is approximately $-840 \, \text{m}$. Since this displacement is negative, it indicates that the projectile ends up 840 meters below its starting point, meaning the height of the cliff is 840 meters.

Would you like any additional details or further clarification?

Here are 5 related questions to expand on this topic:

1. What would be the maximum height reached by the projectile?
2. How would air resistance affect the projectile's motion?
3. Can we calculate the range (horizontal distance) of the projectile?
4. What would happen to the time of flight if the initial vertical velocity were doubled?
5. How does the angle of projection affect the flight of a projectile?

Tip: The horizontal and vertical motions of a projectile are independent of each other. Always treat them separately when solving projectile motion problems.