Calculating Passing Height of Two Thrown Balls

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SUMMARY

The discussion centers on calculating the height at which two vertically thrown balls pass each other. Ball A is thrown with an initial speed of 40 m/s, while Ball B is thrown 1 second later with an initial speed of 47.5 m/s. Using the kinematic equations, specifically s = vt + ½at², participants analyze the motion of both balls to determine the intersection point. The key insight is that both balls will meet at a height determined by their respective velocities and the time elapsed since their launch.

PREREQUISITES
  • Understanding of kinematic equations, specifically s = vt + ½at²
  • Knowledge of vertical motion under gravity, including acceleration due to gravity (g = 10 m/s²)
  • Ability to calculate time of flight for objects in motion
  • Familiarity with concepts of relative motion in physics
NEXT STEPS
  • Calculate the maximum height of Ball A using the equation t = v/g
  • Determine the time of flight for Ball B after its launch
  • Formulate and solve simultaneous equations for the heights of both balls
  • Explore the implications of varying initial speeds on the intersection height
USEFUL FOR

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

Said Ahmad
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Homework Statement



Ball A is thrown vertically by speed 40 m/s. After 1 second, Ball B is thrown vertically also by speed 47.5 m/s.
What height from ground will they pass each other? ( g = 10m/s2 )

Homework Equations



s = vt + ½at2
t = v/g
v2 = v02 + 2as

The Attempt at a Solution



Time A to reach the max height is t = v/g = 40/10 = 4 s

I don't understand. Because there is plus of 1 sec for ball B.
 
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Said Ahmad said:

Homework Statement



Ball A is thrown vertically by speed 40 m/s. After 1 second, Ball B is thrown vertically also by speed 47.5 m/s.
What height from ground will they pass each other? ( g = 10m/s2 )

Homework Equations



s = vt + ½at2
t = v/g
v2 = v02 + 2as

The Attempt at a Solution



Time A to reach the max height is t = v/g = 40/10 = 4 s

I don't understand. Because there is plus of 1 sec for ball B.
What's not to understand?

Both balls are thrown up in the same direction, although at different times and different speeds. The two balls are going to pass each other at some point, either on the way up or on the way down. At what height are the two balls the same distance above the ground?

Hint: write a kinematics equation for each ball.
 

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