If an object is dropped, how fast will it be moving after falling #3 m#?

1 Answer
Apr 11, 2016

Answer:

The velocity after falling 3m would be #58.8 m.s^(-1)#

If drag is ignored.

Explanation:

All objects in free fall near the Earth's surface and with no drag have a constant acceleration of #9.8 m.s^(-2)# *. To answer the question we will ignore the effects of drag acting on the object (as no information about the drag is provided).

State data and select equation:
#s = 3.0 m#
#u = 0# (anything dropped starts with zero velocity)
#v = ?#
#a = 9.8 m.s^(-2)#
#t = ?#

Use #v^2 = u^2 + 2as#

Substitute values into the equation:
#v^2 = u^2 + 2as = 0 + 2 × 9.8 × 3 = 58.8 m.s^(-1)#


The reason for an acceleration of #9.8 m.s^(-2)# is because with no drag the resultant force is equal to the weight only. We can write the following equation:
#F = w = mg#
Where
g is the gravitational field strength. Near the Earth's surface the value of g* is #9.8 N.kg^(-1)#

Newton's second law tells us that resultant force is equal to the product of mass and acceleration:
#F = ma#

Combine the two equations:
#F = mg = ma#
Mass cancels out to leave: #a = g = 9.8 m.s^(-2)#