Calculating Passing Height of Two Thrown Balls

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To determine the height at which two balls pass each other, the problem involves calculating their respective heights over time using kinematic equations. Ball A is thrown first at 40 m/s, reaching its maximum height in 4 seconds, while Ball B is thrown 1 second later at 47.5 m/s. The key is to set up equations for the heights of both balls as functions of time and find the point where these heights are equal. The discussion emphasizes the importance of accounting for the time difference in the throws. Solving these equations will yield the height at which the balls meet.
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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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