What Did I Do Wrong in My Physics Homework Problem?

Click For Summary
SUMMARY

The forum discussion focuses on solving a physics problem involving a diver's motion. The diver springs upward with an initial speed of 1.7 m/s from a 3.0 m board. The correct approach to find the velocity upon striking the water involves using the kinematic equation {V_f}^2 = {V_i}^2 + 2ad, where the displacement is -3.0 m and acceleration is -9.80 m/s². The diver's maximum height above the water can be calculated by combining the height from the board and the height reached during the upward motion.

PREREQUISITES
  • Kinematic equations for motion in one dimension
  • Understanding of initial and final velocity concepts
  • Basic knowledge of acceleration due to gravity (9.80 m/s²)
  • Ability to solve quadratic equations
NEXT STEPS
  • Study the derivation and application of kinematic equations in physics
  • Learn how to analyze projectile motion, including upward and downward trajectories
  • Practice solving problems involving initial velocity and displacement
  • Explore the concept of maximum height in projectile motion scenarios
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in mastering motion analysis in physics problems.

shawonna23
Messages
146
Reaction score
0
Physics Homework Problem--Stuck!

A diver springs upward with an initial speed of 1.7 m/s from a 3.0 m board.

a. Find the velocity with which he strikes the water. (Hint: When the diver reaches the water, his displacement is y = -3.0 m (measured from the board), assuming that the downward direction is chosen as the negative direction.)

b. What is the highest point he reaches above the water?


I tried using this equation, but my answer was wrong. Can someone tell me what I did wrong?

Part a:
d=v*t + 1/2 at^2
-3.0=1.7t + 1/2(-9.80)t^2
4.9t^2 - 1.7t -3.0=0
t= 0.98s

v=d/t
v=3/0.98
v=3.06m/s

I didn't do Part b because Part a was wrong!
 
Physics news on Phys.org
This question was asked before by you and was answered.
 
He's going to travel up in the air, stop at the top, then come back down. This can be described by two equations, one going up and one going down.

For the first part:

[tex]{V_f}^2 = {V_i}^2 + 2ad[/tex]

You know the initial velocity, the final velocity (0), and the acceleration. Solve for the distance traveled.

For the second part you use the same equation, with 3 + part A as your total height. You know the rest of the numbers.
 

Similar threads

Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
12
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
11
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K