Step-by-Step Physics Study Guidance for PHYS 1403 Final Practice Problems

Q1. A stone is projected at a cliff of height h with an initial speed of 42.0 m/s directed at angle 60° above the horizontal. The stone strikes at point A, 5.50 s after launching. Find:

• A. The height h of the cliff

• B. How far is point A horizontally?

Background

Topic: 2D Kinematics & Projectile Motion

This question tests your understanding of projectile motion, specifically how to analyze the vertical and horizontal components of a projectile's trajectory to determine height and range.

Key Terms and Formulas

• Initial velocity ($v_0$)

• Launch angle ($\theta_0$)

• Time of flight ($t$)

• Vertical displacement ($y$)

• Horizontal displacement ($x$)

Key formulas:

• Vertical position: $y = v_0 \sin\theta_0 \cdot t - \frac{1}{2} g t^2$

• Horizontal position: $x = v_0 \cos\theta_0 \cdot t$

Step-by-Step Guidance

1. Break the initial velocity into horizontal and vertical components: $v_{0x} = v_0 \cos\theta_0$ $v_{0y} = v_0 \sin\theta_0$

2. For part A (height): Use the vertical motion equation to solve for $h$ at $t = 5.50$ s: $h = v_{0y} \cdot t - \frac{1}{2} g t^2$

3. For part B (horizontal distance): Use the horizontal motion equation to solve for the distance: $x = v_{0x} \cdot t$

4. Plug in the values for $v_0$, $\theta_0$, and $t$ to calculate $v_{0x}$ and $v_{0y}$.

5. Set up the equations for $h$ and $x$ using the calculated components and time.

Try solving on your own before revealing the answer!

Final Answer:

A. $h = 51.8$ m B. $x = 115.5$ m

We used the kinematic equations for projectile motion, substituting the given values for initial speed, angle, and time to find the height and horizontal distance.