Question

Tom the cat is chasing Jerry the mouse across a table surface 1.2 m off the floor. Jerry steps out of the way at the last second, and Tom slides off the edge at aspeed of 2.4 m/s.

A) Where will Tom strike the floor? The acceleration of gravity is 9.8 m/s^2. Answer in units of m.

B) What speed will Tom have just before he hits? Answer in units of m/s.

Answer 1

First you want to start off with what you already know.
?Y is the change in the y direction which is 1.5 m in the problem. We know that the acceleration in the y direction is 9.81 m/s. We also know that Tom goesoff the table in the horizontal or x direction so his initial velocity in the y direction has to be 0.

?Y = Vit + 1/2at^2 ------- use this kinematics equation since it has the variable you already know
1.5 = (0)(t) + 1/2(9.81)(t^2) -------- solve for t
t ˜ 0.306 s

Now we need to solve for ?X, or the change in the x direction. This will tell you how far from the table he lands. Use this formula: V=?x/t. Since you aresolving for ?x all variables should be components of x. V of x is 6 m/s and t is 0.306s, so ?x must be 1.835 m.

Now you need to find the velocities just before he lands for both x and y directions.
Use the Vf^2 = Vi^2 + 2a?x (?y if you are solving for y components)

Let's do x direction first. Vf, or final velocity is unknown and is what we are solving for. Vi, or initial velocity is 6 m/s. a = 0 for the x direction.?x = 1.835 m.

Vf^2 = (6^2) + 2(0)(1.835)
Vf (x direction) = 6 m/s

Now for the y direction. Vi will be 0 since the initial velocity he had was only in the x direction. a will be 9.81 because gravity will be causing Tom toaccelerate downwards. ?y will be 1.5.

Vf^2 = (0)^2 + 2(9.81)(1.2)
Vf (y direction) = 5.425 m/s

(depending on whether or not the question wants you to use vector quantities, you may or may not place a negative sign in front of the y velocity toindicate it is in the negative direction)