Answer:4 times
Explanation:
Given
Balls are identical and thrown with initial velocity lets say u and 2u because the other ball is thrown with twice the velocity of first
Maximum height reached by first ball is h which is given by
[tex]v^2-u^2=2as[/tex]
[tex]0-u^2=2\times g\times h_1[/tex]
[tex]h_1=\frac{u^2}{2g}[/tex]
For second ball
[tex]v^2-u^2=2as[/tex]
[tex]0-(2u)^2=2\times g\times h_2[/tex]
[tex]h_2=\frac{4u^2}{2g}[/tex]
Thus height gained by second ball will be 4 times of first ball.
