the ball is dropped from top of the tower, after 2 seconds another ball is thrown vertically downward with a speed 40m/sec after how much time and at what distance the ball meets
Share
the ball is dropped from top of the tower, after 2 seconds another ball is thrown vertically downward with a speed 40m/sec after how much time and at what distance the ball meets
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Answer:
let the balls meet at distance h below the top of tower t second after dropping of first ball. The second ball takes time (t−2) seconds.
Explanation:
First ball: h=12gt²…(i)
Second ball: h=40 (t−2)+12g (t−2)² …(ii)
Equating (i) and (ii), we get
40(t−2)+12g (t−2)² = 12 gt²
40(t−2) = 12g [t²−(t−2)²]
40(t−2) = 12×10 (2t−2)×2
4t−8=2t−2
⇒t=3s
h=12gt² = 12×10×3²=45m