Answer
$x = 42.7~m$
Work Step by Step
We can use conservation of energy to find the speed at the end of the ramp:
$U_{g2}+K_2 = U_{g1}+K_1$
$K_2 = U_{g1}-U_{g2}+0$
$\frac{1}{2}mv_2^2 = mgh_1-mgh_2$
$v_2^2 = 2g~(h_1-h_2)$
$v_2 = \sqrt{2g~(h_1-h_2)}$
$v_2 = \sqrt{(2)(9.8~m/s^2)~(25~m-3~m)}$
$v_2 = 20.765~m/s$
According to the geometry, the direction of motion at the end of the ramp is $30^{\circ}$ above the horizontal.
For the next part of the question, we can let $~~v_0= 20.765~m/s$
We can find the time of flight
$y = y_0+v_{0y}t+\frac{1}{2}a_yt^2$
$\frac{1}{2}a_yt^2+v_0~sin(30^{\circ})t+y_0 = 0$
$-4.9~t^2+(20.765)~sin(30^{\circ})t+3 = 0$
$-4.9~t^2+10.38~t+3 = 0$
We can use the quadratic formula:
$t = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$t = \frac{-(10.38)\pm \sqrt{(10.38)^2-(4)(-4.9)(3)}}{(2)(-4.9)}$
$t = -0.26~s, 2.376~s$
We can choose the positive solution. Then $t = 2.376~s$
We can find the horizontal distance traveled in this time:
$x = v_{0x}~t$
$x = v_0~cos~30^{\circ}~t$
$x = (20.765~m/s)~(cos~30^{\circ})~(2.376~s)$
$x = 42.7~m$
