Answer
52.8 m
Work Step by Step
Please see the attached image first.
Let's apply equation 3.6b $V^{2}=u^{2}+2aS$ in the vertical direction to find the vertical displacement of the marble.
$\downarrow V^{2}=u^{2}+2aS$ ; Let's plug known values into this equation.
$V_{y}^{2}=0+2(g)h$
$h=\frac{V_{y}^{2}}{2g}-(1)$
Given that the final velocity makes an angle $65^{\circ}$ with respect to the horizontal. So, from trigonometry, we can get.
$tan65^{\circ}=\frac{V_{y}}{V_{x}}$
$V_{y}=15\space m/s\times 2.14=32.17\space m/s-(2)$
(2)=>(1)
$h=\frac{(32.17\space m/s)^{2}}{2\times9.8\space m/s^{2}}=52.8\space m$
