At what height objects will pass each other?

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Two objects are in motion: one is thrown downward from 60 m with an initial speed of 10 m/s, and the other is propelled upward from ground level at 40 m/s. To find the height at which they pass each other, the equation d = Vi . t + 1/2 a t² is used, with adjustments for their starting heights. The calculated height where they meet is confirmed to be 41 m above the ground. A quicker method involves recognizing their constant relative velocity of 50 m/s to determine the time until they meet.
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Homework Statement



An object is thrown downward with an initial (t = 0) speed of 10 m/s from a height of 60 m above the ground. At the same instant (t = 0), a second object is propelled vertically upward from ground level with a speed of 40 m/s. At what height above the ground will the two objects pass each other?

first object
t0 = 0
V = 10 m/s
H = 60
Second object
t0 = 0
V = 40 m/s
H = 60

Homework Equations


d = Vi . t + 1/2 a t2


The Attempt at a Solution


This equation doesn't see me to work since we need to find at which height objects will pass
 
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hi patelneel1994! :wink:

call the height of either object above the ground "d", and rememb
er that the two objects start at different values of d …

show us what you get :smile:
 
So instead of d I should put h. Find height for each object separately!
 
d or h, it doesn't matter, so long as you use the same symbol for each, but start them at different values :wink:
 
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I got 41 m. Thanks
 
patelneel1994 said:
I got 41 m. Thanks
I confirm that answer. A slightly quicker way, perhaps, is to note that the two objects undergo the same acceleration, so the relative velocity is constant. The closing speed is therefore a constant 10+40=50 m/s. That tells you how long they take to meet.
 
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