Math Problem Statement

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 5.00 ss after it was thrown. What is the speed of the rock just before it reaches the water 27.0 mm below the point where the rock left your hand? Ignore air resistance.

Solution

To solve this problem, let's break it down into steps. We'll use the principles of kinematics to determine the speed of the rock just before it reaches the water.

Given:

  • The total time for the rock to return to the starting point: t=5.00t = 5.00 seconds
  • The distance from the starting point to the water: h=27.0h = 27.0 meters
  • Acceleration due to gravity: g=9.81g = 9.81 m/s²
  • Initial velocity: v0v_0 (to be determined)

Step 1: Find the initial velocity (v0v_0)

When the rock returns to the same height from which it was thrown after 5.00 seconds, the time it takes to reach the highest point (where the velocity is 0) is half of that, i.e., tup=5.002=2.50t_{up} = \frac{5.00}{2} = 2.50 seconds.

The initial velocity can be found using the equation: v0=g×tupv_0 = g \times t_{up}

Step 2: Find the speed of the rock just before it reaches the water

The rock will travel an additional 27.0 meters downward after returning to its original height. To find the speed just before it hits the water, we can use the following kinematic equation: v2=v02+2ghv^2 = v_0^2 + 2gh where:

  • v0v_0 is the velocity of the rock when it reaches the starting point again (which will be the same as the initial velocity due to symmetry),
  • gg is the acceleration due to gravity, and
  • hh is the height.

Let's calculate the initial velocity v0v_0 and then use it to find the final speed vv just before the rock hits the water.The initial velocity of the rock when it was thrown upward is approximately 24.5324.53 m/s.

The speed of the rock just before it reaches the water, 27.0 meters below the point where it was thrown, is approximately 33.6333.63 m/s.

Would you like further details or explanations on any part of the solution? Here are some related questions to consider:

  1. What would happen to the final speed if the height was doubled?
  2. How would air resistance affect the final speed?
  3. Can you derive a general formula for the final speed given any height?
  4. What would the initial velocity need to be for the rock to take 10 seconds to return to the same height?
  5. How would the motion differ if the rock was thrown downward instead of upward?

Tip: When dealing with vertical motion, remember that the initial velocity and the acceleration due to gravity both affect the time it takes to reach the maximum height.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Kinematics
Projectile Motion
Gravity

Formulas

v = u + at
v^2 = u^2 + 2as

Theorems

-

Suitable Grade Level

Advanced High School