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Sagot :
After 0.43 seconds,the balls are at the same height.
Since we are given an initial speed of 35 m/s, and the height of the building which is 15 m.Considering the height to h and after a time of t sec, the balls meet.
The lower ball will travel a distance say h₁ in time t secs
using the formula for the equation of motion,
s= ut+1/2at²
=>h₁=35.t−(1/2)g.t²
while in the same time interval, the other ball will travel a distance of 15-h1
so,
=>15 - h1 = (1/2)g.t²
So now substituting the value of h₁ in the above equation, we get
15−35.t+(1/2)gt² = (1/2)g.t²
=> 15 = 35. t
=>t = 15/35
=>t =0.428 = 0.43 seconds
To know more about speed refer to the link https://brainly.com/question/28224010?referrer=searchResults.
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