Answer:
Explanation:
Given
Two baseballs are fired into a pile of hay such that one has twice the speed of the other.
suppose u is the velocity of first baseball
so velocity of second ball is 2u
suppose [tex]d_1[/tex] and [tex]d_2[/tex] are the penetration by first and second ball
using [tex]v^2-u^2=2 ad[/tex]
where v=final velocity
u=initial velocity
a=acceleration
d=displacement
here v=0 because ball finally stops
[tex]0-u^2=2ad_1----1[/tex]
for second ball
[tex]0-(2u)^2=2ad_2----2[/tex]
divide 1 and 2 we get
[tex]\frac{u^2}{4u^2}=\frac{d_1}{d_2}[/tex]
as deceleration provided by pile will be same
[tex]\frac{1}{4}=\frac{d_1}{d_2}[/tex]
[tex]d_2=4d_1[/tex]
thus faster ball penetrates 4 times of first ball
