How Accurate Are My Initial Velocity and Maximum Height Calculations?

Click For Summary
SUMMARY

The discussion focuses on the calculations of initial velocity, maximum height, and velocity after a specific time for a ball thrown upwards, with a maximum height reached in 1.80 seconds. The correct approach for finding initial velocity involves using the equation v2 = v1 + at, where v2 is the final velocity, a is acceleration due to gravity (-9.8 m/s²), and t is time. For maximum height, the correct formula is s = v1*t + 1/2 at², which requires the accurate initial velocity from part A. The calculations presented initially were incorrect due to misuse of displacement formulas instead of velocity equations.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Knowledge of acceleration due to gravity (-9.8 m/s²)
  • Ability to manipulate algebraic equations
  • Familiarity with concepts of velocity and displacement
NEXT STEPS
  • Review kinematic equations for uniformly accelerated motion
  • Practice solving problems involving initial velocity and maximum height
  • Learn how to differentiate between displacement and velocity calculations
  • Explore real-world applications of projectile motion
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding projectile motion calculations.

bwilhelm
Messages
9
Reaction score
0
Am I anywhere near correct?


Homework Statement


A ball is thrown and upward and reaches it's maximum height in 1.80 sec.

A. Find Initial Velocity
B. Find Max Height Reaced
C. Find Velocity 1.15 sec after being thrown.


Homework Equations





The Attempt at a Solution


A. .5(-9.8)*1.8^2= -15.87 m/sec
B. -15.87+.5(-9.8)*1.8^2 = 31.74m
C. .5(-9.8)1.15^2= -6.48 m/sec
 
Physics news on Phys.org
No, not near correct. For part a, you used a displacement formula... so the result won't be a velocity...

Hint: for part a) use the fact that v2 = v1 + at. You know v2, a and t... use those to calculate v1.

For part b) you made a mistake with the formula... it should be s = v1*t + 1/2 at^2... you didn't multiply v1 by t... and also the v1 is wrong since part a was wrong...

For part c)... here again you use a displacement formula... so the result that comes out will be displacement not a velocity... 1/2 at^2 is a distance not a velocity.
 

Similar threads

Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
Replies
19
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
9
Views
4K
Replies
10
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
5
Views
3K