Math Problem Statement

A hiker throws a stone from the upper edge of a vertical cliff. The stone's initial velocity is 25.0 m/s directed at 40.0° with the face of the cliff, as shown in the figure. The stone hits the ground 3.75 s after being thrown and feels no appreciable air resistance as it falls. What is the height of the cliff?

Solution

To solve this problem, we need to break down the stone's motion into horizontal and vertical components and use kinematic equations to find the height of the cliff.

Step 1: Breakdown of the Initial Velocity

The initial velocity (v0v_0) is 25.0 m/s, directed at an angle of 40.0° with the face of the cliff.

  1. Horizontal component of the velocity: v0x=v0cos(40.0)v_{0x} = v_0 \cdot \cos(40.0^\circ)

  2. Vertical component of the velocity: v0y=v0sin(40.0)v_{0y} = v_0 \cdot \sin(40.0^\circ)

Step 2: Vertical Motion Analysis

The vertical motion of the stone can be described using the kinematic equation: y=v0yt+12at2y = v_{0y} t + \frac{1}{2} a t^2 where:

  • yy is the vertical displacement (which will be negative, as the stone is falling down).
  • v0yv_{0y} is the initial vertical velocity component.
  • t=3.75t = 3.75 s is the time of flight.
  • a=9.8a = -9.8 m/s² is the acceleration due to gravity (negative because it acts downward).

We are interested in finding the height hh of the cliff. The vertical displacement yy from the point of projection to the ground is: y=v0yt+12at2y = v_{0y} t + \frac{1}{2} a t^2

Since the stone falls down from the initial height hh, we have: h=v0yt+12at2-h = v_{0y} t + \frac{1}{2} a t^2 So, the height of the cliff hh can be calculated as: h=(v0yt+12at2)h = -\left( v_{0y} t + \frac{1}{2} a t^2 \right)

Step 3: Calculation

Let's calculate the height step by step.

  1. Calculate the vertical component of the initial velocity: v0y=25.0m/ssin(40.0)v_{0y} = 25.0 \, \text{m/s} \cdot \sin(40.0^\circ)

  2. Use the kinematic equation to find hh: h=(v0y3.75s+12(9.8m/s2)(3.75s)2)h = -\left( v_{0y} \cdot 3.75 \, \text{s} + \frac{1}{2} \cdot (-9.8 \, \text{m/s}^2) \cdot (3.75 \, \text{s})^2 \right)

Let's compute this.The height of the cliff is approximately 8.64 meters.

Would you like more details or have any further questions?

Here are 5 related questions you might find interesting:

  1. How far horizontally did the stone travel before hitting the ground?
  2. What was the final velocity of the stone just before impact?
  3. How would the height change if the angle was increased to 50°?
  4. What would be the time of flight if the stone was thrown vertically downward?
  5. How would air resistance affect the height calculation?

Tip: Breaking down vectors into their components simplifies solving problems involving angles.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Projectile Motion
Kinematic Equations
Vectors

Formulas

Horizontal and Vertical Components of Velocity
Kinematic Equation for Vertical Motion

Theorems

-

Suitable Grade Level

Grades 11-12