Max Height & Time In Air of Ball Thrown Upward from 25m

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SUMMARY

A ball is thrown straight up from a height of 25 meters with an initial speed of 11 m/s. The maximum height reached by the ball is calculated to be 31 meters. To determine the total time the ball is in the air, one must first calculate the time taken to reach the maximum height using the equation v = u + at, where 'v' is the final velocity (0 m/s at the peak), 'u' is the initial velocity (11 m/s), and 'a' is the acceleration due to gravity (-9.8 m/s²). The total time in the air is then found by doubling the time to reach the maximum height and adding the time taken to descend back to the original height of 25 meters.

PREREQUISITES
  • Understanding of kinematic equations, specifically s = ut + 0.5at²
  • Familiarity with the concept of acceleration due to gravity (9.8 m/s²)
  • Knowledge of initial velocity and its role in projectile motion
  • Ability to solve quadratic equations
NEXT STEPS
  • Study the kinematic equation s = ut + 0.5at² in detail
  • Learn how to derive maximum height in projectile motion problems
  • Explore the concept of free fall and its equations
  • Investigate real-world applications of projectile motion in sports and engineering
USEFUL FOR

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

sp00ky
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A ball is thrown straigh up, from height of 25 m above the ground, at a speed of 11 m/s.
The first question is what is the max. height of the ball above the ground.
I got that one it's like 31m. But there was another question after that that asks how long is the ball in the air. What equation must I use to find this out?
 
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When does the ball 'stop' being in the air?
 
well first find the time taken to reach the max ht. double this time it is the time taken for the ball to reach the ht it was thrown from(25m ) at this pt the ball has a speed of 11m/s use s =ut+4.9t^2 to find this t2 add this to t1 and that is the time taken to reach the ground. to find time to reach max ht use v = u+at.
 

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