Solving Kinematics: A Ball Thrown Upward

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SUMMARY

A ball thrown vertically upward at a speed of 25.2 m/s reaches a maximum height of 32.37 meters and takes 2.57 seconds to reach this peak. The time taken for the ball to return to the ground after reaching its highest point is also 2.57 seconds, confirming the symmetry of projectile motion. The velocity of the ball upon returning to the original launch level is equal to the initial velocity of 25.2 m/s, as per the principles of kinematics.

PREREQUISITES
  • Understanding of kinematics equations, particularly the equation x = x0 + v0t + (1/2)at^2.
  • Knowledge of projectile motion and its symmetrical properties.
  • Familiarity with concepts of velocity and acceleration due to gravity.
  • Basic algebra skills for solving equations.
NEXT STEPS
  • Study the derivation and application of kinematics equations in various scenarios.
  • Explore the effects of different initial velocities on the maximum height and time of flight.
  • Learn about the concept of free fall and its relation to gravitational acceleration.
  • Investigate real-world applications of projectile motion in sports and engineering.
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Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of projectile motion and its calculations.

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Homework Statement



A ball is thrown vertically upward with a speed of 25.2 m/s.
a)How high does it rise? Answer in units of m.
b) How long does it take to reach its highest point? Answer in units of s.
c) How long does it take the ball to hit the ground after it reaches its highest point? Answer in units of s.
d) What is its velocity when it returns to the level from which it started? Answer in units
of m/s.

Homework Equations


all kinematics equations.
I used x=x0 + v0t + (1/2)at^2, I remember.



The Attempt at a Solution


I have a thru c - a was 32.37, and b abd c were both 2.57 (parabola was symmetrical) but I have no idea how to solve d, at all.
 
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Since the ball is taking the same time to rise as it is to fall again, this would mean that by the time the ball reached the same level it was as at when it started, it would be going the same speed as it was when it was first thrown upwards.
 

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