If an object is dropped, how fast will it be moving after falling #25 m#?

1 Answer
Feb 4, 2016

Answer:

Assuming no air resistance, it will be moving at #v=sqrt(2ad) = sqrt(2*9.8*25)=sqrt(490)=22.1# #ms^-1#

Explanation:

The expression for how fast an object will be travelling after covering a certain distance while accelerating is:

#v^2=u^2+2ad#

Where:

#v# = final velocity #(ms^-1)#
#u# = initial velocity #(ms^-1)#
#a# = acceleration #(ms^-2)#
#d# = distance #(m)#

In this case, we assume that the initial velocity is #0#: the question says the object was 'dropped', not 'thrown'.

We can leave out the #u# term and take the square root of both sides to make #v# the subject:

#v=sqrt(2ad)#

Substituting in our values, including the acceleration due to gravity:

#v = sqrt(2*9.8*25)=sqrt(490)=22.1# #ms^-1#

We have assumed that there is no air resistance acting, or that we can neglect it because it is much smaller than the gravitational force.