Question

An egg is dropped from a height of 20 m. The egg is in free fall the entire time it falls. What wilLthe speed of the egg be when it lands after fallinG for 2 seconds?

Answer 1

This involves a kinematics equation for constant acceleration, which contains the variables of velocity initial, velocity final, time and acceleration: V final = V initial + AT
Acceleration = -9.8
V initial = 0
V final = ?
Time = 2
V final = 0 + (-9.8)(2)
V final = -19.6 m/s

Answer 2

The speed of the egg when it lands after falling for 2 seconds is 20 meter per seconds.

Given the following data:

• Height = 20 meters

• Initial velocity = 0 m/s(since the egg is starting from rest).

• Time = 2 seconds

We know that the acceleration due to gravity (a) for an object in free fall is equal to 10 meter per seconds square.

To find the speed of the egg when it lands after falling for 2 seconds, we would use the first equation of motion;

$V = U + at$

Where:

• U is the initial velocity.

• V is the final velocity.

• a is the acceleration.

• t is the time measured in seconds.

Substituting the given parameters into the formula, we have;

$V = 0 + 10(2)\\\\V = 0 + 20$

Final velocity, V = 20 m/s.

Therefore, the speed of the egg when it lands after falling for 2 seconds is 20 meter per seconds.

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