## College Physics (4th Edition)

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