First, let's determine the x and y-components of the initial velocity. \[v_x = v \cos(37) ~~{\rm and}~~ v_y = v \sin(\theta)\]If we are to ignore wind-resistance, we know that \(v_x\) will remain constant throughout the the entire flight since no forces act in the x-direction. We know that \(v_y\) will initially decrease as the projectile heads upwards. Once the projectile reaches the top of its trajectory (\(v_y=0\) here) it will accelerate towards the earth.
Now, we need to determine how high the object is at the top of it's trajectory. We will use the launch point as the reference for this portion. From conservation of energy\[\Delta KE = \Delta PE\]which means that all the kinetic energy that the projectile possesses at launch will all be converted to potential energy when the projectile reaches the top of its trajectory. \[{1 \over 2} m (v_y)^2 = mg h \rightarrow {1 \over 2} v_y^2 = gh\]we know \(v_y\) and g, therefore we can solve for h. Now, remember this is the height that the projectile gained from its initial launch point. The launch point is 27m above the surface of the earth.
Therefore, now we need to consider the potential energy of the projectile from the surface of the earth. Again, we can use conservation of energy \[{1 \over 2} v_{y,i}^2 = g(h+27)\]where h is the height we calculated previously and \(v_{y,i}\) is the velocity in the y-direction at the time of impact. We can solve for \(v_{y,i}\).
Now, to determine the magnitude of velocity, we need to use Pythagorean's Theorem\[v_i = \sqrt{~v_x^2 + v_{y,i}^2}\]
For part b, the vertical component of velocity can be found as\[{1 \over 2} v_{y,i}^2 = {1 \over 2} v_y^2 + gh \]where \(v_{y,i}\) is the impact velocity in the y-direction, \(v_y\) is the initial velocity, and h is the height of the cliff.