If a egg is dropped from eleven meters how fast will it be going?

Answer 1

v^2 -u^2 = 2 * a * s v=final velocity u=initial velocity a=acceleration s=displacement(distance) as u equals zero v = sqrt(2 * a * s) a=9.8 (acceleration due to gravity on Earth's surface m/s/s) s=11 v = sqrt(2*9.8*11) v = sqrt(215.6) v = 14.7 m/s

Answer 2

The speed at which the egg will be traveling when it hits the ground depends on various factors such as air resistance and initial velocity. Assuming the egg is in free fall and neglecting air resistance, its speed when it hits the ground can be calculated using the equation (v = \sqrt{2 \cdot g \cdot h}), where (v) is the final velocity, (g) is the acceleration due to gravity (approximately 9.81 m/s²), and (h) is the height (11 meters in this case). This would give a final velocity of approximately 14 m/s.