Velocity/Speed Problems (Studying for exam)

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SUMMARY

The discussion focuses on a physics problem involving the calculation of the maximum height and velocity of a ball thrown vertically with an initial speed of 13.5 m/s. The maximum height calculated is 9.42 meters, achieved after 1.38 seconds. Additionally, the discussion highlights the need for a kinematic equation to determine the velocity of the ball at a height of 1.50 meters above the starting point. The correct application of kinematic equations is emphasized for accurate results.

PREREQUISITES
  • Understanding of kinematic equations in physics
  • Basic knowledge of projectile motion
  • Familiarity with acceleration due to gravity (9.81 m/s²)
  • Ability to perform calculations involving time, velocity, and displacement
NEXT STEPS
  • Learn the kinematic equation for final velocity: V² = V₀² + 2a(x - x₀)
  • Study the concept of maximum height in projectile motion
  • Explore the implications of air resistance on projectile motion calculations
  • Practice additional problems involving vertical motion and kinematic equations
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Students studying physics, educators teaching kinematics, and anyone preparing for exams in mechanics or related fields.

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Homework Statement


A person throws a ball with 13,5 m/s straight up. Calculate the maximum height that it will reach from the point where it was thrown. Also find the velocity when the ball is 1.50m above the starting point.

Sorry if the problem is hard to understand, had to translate it from spanish, hopefully it's clear enough.

Homework Equations


Vx = V0x + axt
x = x0 + v0xt + 1/2 ax t^2

The Attempt at a Solution


13.5m/s = 0 + 9.81 * t
13.5/9.81 = t
t = 1.38 s (what it took to reach the highest point)

x = 0 + 13.5 * 1.38 + 1/2 * -9.81 * 1.37^2
x = 9.42 m (maximum height?)

Not sure if the problem is correct, if it's not any guidance is appreciated. I'm also missing what's the velocity 1.50m above the starting point.
 
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t is correct (1.376 s), but you are using 1.38 and 1.37^2 in the same equation. Hang on to the couple extra digits until you have the final result.

You know x0, x, Vx0, and a. There is a kinematic equation that you have not posted that will allow you to find vx directly.
 

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