Tom the cat is chasing Jerry the mouse across a table surface 1.5 m off the floor. Jerry steps out of the way at the last second, and Tom slides off the edge of the table at a speed of 6.0 m/s. Where will Tom strike the floor?
_______m from the table
What velocity components will he have just before he hits? (Use a coordinate system in which up is positive.)
_______m/s (x direction)
_______m/s (y direction)
_______m from the table
What velocity components will he have just before he hits? (Use a coordinate system in which up is positive.)
_______m/s (x direction)
_______m/s (y direction)
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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 goes off 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 are solving 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 to accelerate 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 to indicate it is in the negative direction)
∆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 goes off 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 are solving 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 to accelerate 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 to indicate it is in the negative direction)