How High Does a Ball Go When Thrown Upward at 25m/s?

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SUMMARY

A ball thrown straight up at an initial velocity of 25 m/s reaches a maximum height determined by the kinematic equations of motion. The relevant equations include S = v*t and V = v₀ + at, where gravity is represented as -9.81 m/s². To find the maximum height, one must first calculate the time to reach the peak using the equation T = v₀/g, where v₀ is the initial velocity and g is the acceleration due to gravity. The maximum height can then be calculated using the equation S = v₀*T - 0.5*g*T².

PREREQUISITES
  • Understanding of kinematic equations for accelerated motion
  • Basic knowledge of physics concepts such as velocity and acceleration
  • Familiarity with the concept of gravitational acceleration (-9.81 m/s²)
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Learn how to derive the maximum height using kinematic equations
  • Explore the concept of projectile motion and its equations
  • Study the effects of air resistance on projectile motion
  • Investigate real-world applications of kinematic equations in sports and engineering
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of motion under gravity.

JBemp
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Homework Statement


A ball is thrown stright up with the speed 25m/s
how high does the ball go?

Homework Equations


S=v*t or V=volt+at
I know gravity is a neg acceloration so I should have -9.81 somewhere



The Attempt at a Solution


I need to find the time first am guessing so, T=s/v but i don't know S ether just 25m/s and -9.81 I know this is most likly very simple but it just eludes me
 
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