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a ball is thrown upward from the ground with an initial speed of 35 m/s; at the same instant, another ball is dropped from a building 15 m high. after how long will the balls be at the same height?



Answer :

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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