IDNLearn.com is designed to help you find reliable answers quickly and easily. Get prompt and accurate answers to your questions from our experts who are always ready to help.
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.
#SPJ4
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.
