01 - Highest point of projectile as measured from inclined plane

$v_o = 15 \, \text{ m/s}$

$v_{ox} = 15 \cos 60^\circ = 7.5 \, \text{ m/s}$

$v_{oy} = 15 \sin 60^\circ = 12.99 \, \text{ m/s}$
 

$v_{Ax} = v_{ox} = 7.5 \, \text{ m/s}$

$v_{Ay} = 7.5 \tan 30^\circ = 4.33 \, \text{ m/s}$
 

 

$v_{Ay} = v_{oy} - gt$

$4.33 = 12.99 - 9.81t$

$t = 0.8828 \, \text{ s}$
 

$x = v_{ox}t = 7.5(0.8828)$

$x = 6.62 \, \text{ m}$
 

${v_{Ay}}^2 = {v_{oy}}^2 - 2gy$

$4.33^2 = 12.99^2 - 2(9.81)y$

$y = 7.64 \, \text{ m}$
 

$a = x \tan 30^\circ = 6.62 \tan 30^\circ$

$a = 3.82 \, \text{ m}$
 

$\cos 30^\circ = \dfrac{z}{y - a}$

$z = (y - a) \cos 30^\circ$

$z = (7.64 - 3.82) \cos 30^\circ$

$z = 3.31 \, \text{ m}$           answer