Answer:
The height from which the egg is dropped is 68.54 m.
Explanation:
Given:
Initial velocity of egg is, [tex]u=0\ m/s[/tex](Dropped means initial velocity is 0)
Time taken is, [tex]t=3.74\ s[/tex]
Acceleration experienced by egg is due to gravity, [tex]a=g=9.8\ m/s^2[/tex]
The height from which the egg is dropped is, [tex]d=?[/tex]
Now, we use Newton's equation of motion that relates the distance, initial velocity, time and acceleration.
So, we have the following equation of motion:
[tex]d=ut+\frac{1}{2}at^2[/tex]
Plug in all the given values and solve for 'd'. This gives,
[tex]d=0+\frac{1}{2}\times 9.8\times (3.74)^2\\\\d=\frac{1\times 9.8\times 13.9876}{2}\\\\d=\frac{137.078}{2}\\\\d=68.54\ m[/tex]
Therefore, the height from which the egg is dropped is 68.54 m.
