#### Answer

Julia should drop the paper a horizontal distance of $35.5~m$ before the driveway.

#### Work Step by Step

We can find the time it takes the paper to fall 1.00 meter:
$y = \frac{1}{2}a_yt^2$
$t = \sqrt{\frac{2y}{a_y}}$
$t = \sqrt{\frac{(2)(1.00~m)}{9.80~m/s^2}}$
$t = 0.452~s$
We can find the horizontal distance the paper moves in this time:
$d = v~t = (15~m/s)(0.452~s) = 6.78~m$
We can find the paper's rate of deceleration as it slides along the ground:
$mg~\mu_k = ma$
$a = g~\mu_k$
$a = (9.80~m/s^2)(0.40)$
$a = 3.92~m/s^2$
We can find the distance the paper slides on the ground until it stops:
$v_f^2 = v_0^2+2ax$
$x = \frac{v_f^2 - v_0^2}{2a}$
$x = \frac{0 - (15~m/s)^2}{(2)(-3.92~m/s^2)}$
$x = 28.7~m$
The total horizontal distance the paper travels after it is dropped is $6.78~m+28.7~m$ which is $35.5~m$
Julia should drop the paper a horizontal distance of $35.5~m$ before the driveway.