To solve this, we need to determine the maximum height the projectile reaches above the cliff top. The maximum height occurs when the vertical component of the velocity becomes zero. We can use kinematic equations for projectile motion to achieve this.

Step 1: Decompose the initial velocity

The initial velocity is $v_0 = 65.0 \, \text{m/s}$, and the launch angle is $\theta = 35.0^\circ$.

• The vertical component of the initial velocity is: $v_{0y} = v_0 \sin(\theta) = 65.0 \sin(35.0^\circ)$

• The horizontal component of the initial velocity (though it's not required for finding height) is: $v_{0x} = v_0 \cos(\theta) = 65.0 \cos(35.0^\circ)$

Step 2: Use the vertical kinematic equation

We use the following kinematic equation for the vertical motion to find the maximum height:

$v_y^2 = v_{0y}^2 - 2g h_{\text{max}}$

At the maximum height, the vertical velocity $v_y = 0$, so the equation becomes:

$0 = v_{0y}^2 - 2g h_{\text{max}}$

Solving for $h_{\text{max}}$:

$h_{\text{max}} = \frac{v_{0y}^2}{2g}$

where $g = 9.8 \, \text{m/s}^2$.

Step 3: Calculate the maximum height

Now, let’s plug in the values and solve for $h_{\text{max}}$. I will calculate that next.The maximum height reached by the projectile above the cliff top is approximately $70.92 \, \text{m}$.

Would you like further details or have any questions about this calculation?

Here are 5 related questions to consider:

1. How long does it take for the projectile to reach its maximum height?
2. What is the total time the projectile remains in the air?
3. What is the horizontal distance traveled by the projectile before it hits the ground?
4. What is the speed of the projectile just before it hits the ground?
5. How does the angle of launch affect the range and height of a projectile?

Tip: For projectile motion, always decompose the velocity into horizontal and vertical components to simplify the calculations.