The ball's horizontal position [tex]x[/tex] at time [tex]t[/tex] is
[tex]x=v_{0x}t[/tex]
The ball has initial velocity in the horizontal direction only, so [tex]v_{0x}=15\,\frac{\mathrm m}{\mathrm s}[/tex]. Then the time it takes for the ball to travel 83 m horizontally [tex]t[/tex] is given by
[tex]83\,\mathrm m=\left(15\,\dfrac{\mathrm m}{\mathrm s}\right)t\implies t=5.5\,\mathrm s[/tex]
Meanwhile, the ball's vertical position [tex]y[/tex] at time [tex]t[/tex], starting at the height of the cliff [tex]h[/tex], is given by
[tex]y=h-\dfrac12gt^2[/tex]
where [tex]g=9.8\,\frac{\mathrm m}{\mathrm s^2}[/tex] is the acceleration due to gravity. After 5.5 seconds, the ball hits the ground, so that [tex]y=0[/tex] when [tex]t=5.5\,\mathrm s[/tex], and we use this to solve for [tex]h[/tex]:
[tex]0=h-\dfrac12g(5.5\,\mathrm s)^2\implies h\approx150\,\mathrm m[/tex]
