Question

Question #91249

Answer 1

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`