Marcus Aurelius is playing with his mouse cat toy. He throws the mouse toy straight up in the air with an initial velocity of 3.5 m/s. How long (how many seconds) until the mouse toy returns to him? Air resistance is negligible.

Hint: There are multiple ways to solve this problem. Some things to consider while solving this problem...

What is the acceleration of the toy?
What will be the displacement when the toy returns to him?
What will be the final velocity at the top of the trajectory?
What will be the final velocity when Marcus catches the toy?

1 Answer
devi f.
Feb 27, 2018

See below, I will show the concepts. You do the data calculation!!

Explanation:

Recall the 3 equations of motion,

  1. Relates time and position
  2. Relates time and velocity.
  3. Relates position and velocity

You need to select the one which relates velocity and time, as you know the initial velocity of the throw.

So initial velocity =3.5m/s
When it reaches top of its trajectory and about to start falling its velocity will be zero.
So: Final velocity for one-half of the throw=0m/s

Solve equation 2:

#v=u +at#
where #v=0#
#u = 3.5#m/s
#a=-9.81#m/#sec^2#

Solving will give you the time it took to reach the peak of its height.

Double that and you have the total time.

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