Answer
The skater should start this stunt a distance of 15.8 meters from the bottom of the ramp.
Work Step by Step
We can find the time it takes to fall 3.00 meters:
$\Delta y = \frac{1}{2}a_y~t^2$
$t = \sqrt{\frac{2\Delta y}{a_y}}$
$t = \sqrt{\frac{(2)(3.00~m)}{9.80~m/s^2}}$
$t = 0.7825~s$
We can find the required horizontal velocity to travel 7.00 meters in this time:
$v_x = \frac{7.00~m}{0.7825~s} = 8.946~m/s$
We can find the acceleration while moving down the ramp:
$mg~sin~\theta = ma$
$a = g~sin~\theta$
$a = (9.80~m/s^2)~sin~15.0^{\circ}$
$a = 2.536~m/s^2$
We can find the distance the skater needs to travel down the ramp in order to have a velocity of $8.946~m/s$ at the bottom:
$v_f^2 = v_0^2+2ad$
$d = \frac{v_f^2-v_0^2}{2a}$
$d = \frac{(8.946~m/s)^2-0}{(2)(2.536~m/s^2)}$
$d = 15.8~m$
The skater should start this stunt a distance of 15.8 meters from the bottom of the ramp.
