Question #91249

1 Answer
Jan 27, 2017

Use this equation of motion to solve problems in which you do not know the final velocity

#Deltay=v_oDeltat+1/2a(Deltat)^2#

I'll show you how below.

Explanation:

The equation above has been derived especially for the purpose of solving a problem in which you do not know the final velocity of the object.

#Deltay# is the distance the object will rise or fall during a given time interval.

#v_o# is the starting velocity of the object

#Deltat# is the time interval beginning from when the velocity was #v_o#. (It is zero in your problem.)

#a# is the (constant) acceleration of the object during the period #Deltat#. (It is -10 in your problem. The - sign means it acts downward.)

The first term in the equation tells us how far the object would travel in time #Deltat# if it were to remain at velocity #v_o# and not accelerate.

The second term tells us the additional distance it travels due to the acceleration. (It is the complicated part!)

So, since in your problem, #v_o=0#, all we do is insert the time (1.9 s) and solve for #Deltay#

#Deltay=0-1/2(10)(1.9)^2= -18.1 m#

The negative sign means the object is travelling downward.

In the first 5 seconds, it falls

#Deltay=0-1/2(10)(5)^2= -125 m#