Question

Two identical balls are thrown vertically upward. The second ball is thrown with an initial speed that is twice that of the first ball. How does the maximum height of the two balls compare?

Answer 1

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

$v^2-u^2=2as$

$0-u^2=2\times g\times h_1$

$h_1=\frac{u^2}{2g}$

For second ball

$v^2-u^2=2as$

$0-(2u)^2=2\times g\times h_2$

$h_2=\frac{4u^2}{2g}$

Thus height gained by second ball will be 4 times of first ball.