Question

Wo baseballs are fired into a pile of hay. If one has twice the speed of the other, how much farther does the faster baseball penetrate?

Answer 1

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 $d_1$ and $d_2$ are the penetration by first and second ball

using $v^2-u^2=2 ad$

where v=final velocity

u=initial velocity

a=acceleration

d=displacement

here v=0 because ball finally stops

$0-u^2=2ad_1----1$

for second ball

$0-(2u)^2=2ad_2----2$

divide 1 and 2 we get

$\frac{u^2}{4u^2}=\frac{d_1}{d_2}$

as deceleration provided by pile will be same

$\frac{1}{4}=\frac{d_1}{d_2}$

$d_2=4d_1$

thus faster ball penetrates 4 times of first ball