Motion problem? (Ball is thrown upward)

  • Thread starter Thread starter toyotadude
  • Start date Start date
  • Tags Tags
    Motion
Click For Summary
SUMMARY

The discussion centers on solving a motion problem involving a ball thrown upward, where the initial velocity (V(initial)) is 0 m/s and the acceleration due to gravity (a) is -9.81 m/s². The participant initially assumed equal time for ascent and descent, calculating time (t) as 0.84 seconds. However, they incorrectly applied the formula "Distance = Speed x Time" without accounting for acceleration, leading to an erroneous solution. The correct approach emphasizes using formulas that incorporate acceleration for accurate results.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Familiarity with concepts of acceleration and gravity
  • Knowledge of how to apply the formula for displacement
  • Basic algebra skills for solving equations
NEXT STEPS
  • Study kinematic equations for uniformly accelerated motion
  • Learn how to derive displacement using initial and final velocities
  • Explore the concept of free fall and its implications in motion problems
  • Practice solving similar motion problems with varying initial conditions
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone seeking to improve their problem-solving skills in motion-related scenarios.

toyotadude
Messages
18
Reaction score
0

Homework Statement



r1x8hx.jpg


The Attempt at a Solution



I assumed that it was the same time going up as it was going down, and thus I had:

V(initial) = 0 m/s
a = -9.81 m/s^2
t = .84 seconds

And to solve for speed, I used the Distance = Speed x Time, thus Speed = Distance/Time... and used the Displacement = 1/2(V(initial) + V(final))*time to find the "distance", and then divide it by .84 seconds... but this was incorrect? ):
 
Physics news on Phys.org
"Distance = Speed x Time" does not apply because the motion is accelerated. Only use formulas with acceleration in them. Your insight that the time to fall is equal to the time to rise may well simplify the problem - consider only the falling part.
 

Similar threads

Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
5K
  • · Replies 38 ·
2
Replies
38
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
Replies
3
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K