Answer
(a) $x=0.66 m$
(b) $y=0.206m$
Work Step by Step
(a)
First, we find the initial velocity.
$v_{\circ}=\sqrt{2gh}$
$\implies v_{\circ x}=\sqrt{2(9.8m/s^2)(0.39m-0.11m)}=2.34m/s$
We know that
$y=y_{\circ}+v_{\circ}t+\frac{1}{2}a_yt^2$
We plug in the known values to obtain:
$0=0.11m+(2.34m/s)sin 36^{\circ}t-\frac{1}{2}(9.8m/s^2)t^2$
This simplifies to:
$t=0.346s$
Now $x=v_{\circ x}t$
We plug in the known values to obtain:
$x=(2.34m/s)(\cos 36^{\circ})(0.346s)$
$x=0.66 m$
(b) We know that
$v_y=v_{\circ}sin\theta-gt$
$\implies t=\frac{v_{\circ}sin\theta}{g}$
$\implies t=\frac{(2.34m/s)sin36^{\circ}}{9.8m/s^2}$
$\implies t=0.140s$
Now the maximum height can be determined as
$y_{max}=y_{\circ}+v_{\circ y}t+\frac{1}{2}a_y t^2$
We plug in the known values to obtain:
$y_{max}=0.11m+(2.34m/s)sin 36^{\circ}(0.140s)-\frac{1}{2}(9.8m/s^2)(0.140s)^2$
$\implies y_{max}=0.206m$