Answer
(a) The initial speed of the egg is $18.5\frac{m}{s}$.
(b) The egg reaches a maximum height of $17.5m$ from its starting point.
(c) The egg is momentarily at rest at the highest point of its motion. So the magnitude of its velocity at the highest point of its motion is 0.
(d) The magnitude and direction of the acceleration at the highest point is $9.80\frac{m}{s^2}$ and downwards.
(e) See the graph below.
Work Step by Step
Let us take the point $30.0m$ below the egg's initial position as the origin of our system with (+)ve y-direction upwards.
Since the motion is free fall motion, we can use the kinematics equations.
(a) Let us calculate the initial speed of the egg.
$y-y_{0}=v_{0y}t+\frac{1}{2}a_{y}t^2$
$=>v_{0y}=\frac{y-y_{0}}{t}-\frac{1}{2}a_{y}t=\frac{-30.0}{5.00}+\frac{1}{2}\times9.80\times5.00=18.5\frac{m}{s}$
The initial speed of the egg is $18.5\frac{m}{s}$.
(b) Now, let us calculate the maximum height the egg reached above the starting point.
$2a_{y}(y-y_{0})=v_{y}^2-v_{0y}^2$
$=>s=\frac{-v_{0y}^2}{2a_{y}}$ ,$[s=y-y_{0}=y-30.0]$
$=>s=\frac{-18.5^2}{-2\times9.80}=17.5m$
The egg reaches a maximum height of $17.5m$ from its starting point.
(c) The egg is momentarily at rest at the highest point of its motion. So the magnitude of its velocity at the highest point of its motion is 0.
(d) The motion of the egg is free-fall motion. So, the acceleration is constant and its direction is downwards throughout the motion.
Therefore, the magnitude and direction of the acceleration at the highest point is $9.80\frac{m}{s^2}$ and downwards.
(e) See the graph.
