Question

A ball is dropped from the top of tower of height 78.4m another ball is thrown down with a certain velocity 2sec later.If both the balls reaches the ground simultaneously. the velocity of the second ball is
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Answer 1

Question:

A ball is dropped from the top of tower of height 78.4m another ball is thrown down with a certain velocity 2sec later.If both the balls reaches the ground simultaneously. the velocity of the second ball is

​Answer:

velocity of first ball = 39.2

velocity of second ball = 39.25m/s

Explanation:

Given =>

For the 1st ball = Height(S) = 78.4m

                           Initial speed(u) = 0m/s  (ball has being drooped)

                           accleration (a) = 9.8m/s  

                            Final velocity(v) = ?

By the 3rd equation of motion

v^2 = u^2 + 2aS

v^2 = 0^2 + 2 * 9.8 * 78.4

v^2 = 19.6 * 78.4

v^2 = 1536.64

v = √1536.64

v = 39.2 ---------- velocity of first ball

Similarly , for second ball

Given => Height(S) = 78.4m

               Initial speed(u) = 2m/s (ball is being thrown)

               accleration (a) = 9.8m/s  

               Final velocity(v) = ?

By the 3rd equation of motion

v^2 = u^2 + 2aS

v^2 = 2^2 + 2 * 9.8 * 78.4

v^2 = 4 + 1536.64

v^2 = 1540.64

v = √1540.64

v = 39.25m/s ------------- velocity of second ball

So,

velocity of first ball = 39.2

velocity of second ball = 39.25m/s

Hope you understand.