Question

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

Answer 1

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