Answer
The lump hits the wall at a height of $~~2.64~m$
Work Step by Step
We can find the speed of the lump when it leaves the wheel:
$v_0 = \frac{2\pi~r}{T}$
$v_0 = \frac{(2\pi)~(0.200~m)}{5.00\times 10^{-3}~s}$
$v_0 = 251.3~m/s$
If the lump leaves the wheel at the five o'clock position, the angle of the initial velocity above the horizontal is $30^{\circ}$
We can find the time to reach the wall:
$x = v_x~t$
$t = \frac{x}{v_x}$
$t = \frac{2.50~m}{(251.3~m/s)~cos~30^{\circ}}$
$t = 0.01149~s$
We can find the height of the lump when it hits the wall:
$y = y_0+v_{0y}~t+\frac{1}{2}a_y~t^2$
$y = (1.20~m)+(251.3~m/s)(sin~30^{\circ})(0.01149~s)+\frac{1}{2}(-9.8~m/s^2)~(0.01149~s)^2$
$y = 2.64~m$
The lump hits the wall at a height of $~~2.64~m$
