Tom the cat is chasing Jerry the mouse across the surface of a table 1.5 m above 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 5 m/s. What is the Y of Tom's velocity when his velocity vector is pointing down at an angle of 45 degrees
I found that the magnitude and direction of velocity right before Tom hits the floor is:
x=2.75m
vy=-5.39m/s
thanks in advance
tanA = Y/X,
tan45 = Y/5,
Y = 5tan45 = 5m/s.
Well, well, well, look at Tom go! He sure knows how to make an exit, doesn't he?
Now, when Tom's velocity vector is pointing down at an angle of 45 degrees, it means that he's taking a dive towards the floor in stylish fashion. So, let's see what his Y velocity will be in this epic moment.
You've already done the math and found that Tom's vy (vertical velocity) right before hitting the floor is -5.39 m/s. This negative sign indicates that he's indeed going down, not up. Good job on getting that right!
So, with a drumroll, I present to you the Y velocity of Tom when his velocity vector is pointing down at an angle of 45 degrees: -5.39 m/s.
Tom's got moves, even when he's falling! Keep enjoying his adventures with Jerry, and feel free to ask if you need any more humorous assistance!
To find the y-component of Tom's velocity when his velocity vector is pointing down at an angle of 45 degrees, we can first determine the magnitude of the y-component of his velocity.
Given that the velocity in the x-direction (vx) is 2.75 m/s and the velocity in the y-direction (vy) is -5.39 m/s (negative because it is pointing downwards), we can use the Pythagorean theorem to find the magnitude of Tom's velocity before hitting the floor.
Using the formula:
magnitude of velocity = √(vx^2 + vy^2)
We can substitute the given values:
magnitude of velocity = √(2.75^2 + (-5.39)^2)
= √(7.5625 + 29.0521)
= √(36.6146)
= 6.05 m/s (approx.)
Now, to find the y-component of Tom's velocity when his velocity vector is pointing down at an angle of 45 degrees, we can use trigonometry.
The velocity vector is pointed downwards at an angle of 45 degrees, which forms a right angle triangle with the y-component of Tom's velocity and the magnitude of the velocity. Since we have the magnitude and the angle, we can use the sine function to find the y-component.
Using the formula:
y-component of velocity = magnitude of velocity * sin(angle)
Substituting the values:
y-component of velocity = 6.05 m/s * sin(45 degrees)
= 6.05 m/s * 0.7071 (approx.)
= 4.28 m/s (approx.)
Therefore, the y-component of Tom's velocity when his velocity vector is pointing down at an angle of 45 degrees is approximately 4.28 m/s.
To find the y-velocity (vy) when Tom's velocity vector is pointing down at an angle of 45 degrees, we can use the concept of vector components.
The given value of Tom's velocity right before hitting the floor is vy = -5.39 m/s. We need to find the y-component of this velocity when the angle is 45 degrees.
Step 1: Decompose the given velocity into its x and y-components.
The velocity vector forms a right-angled triangle with the x and y-axes. The given velocity is already in the y-direction, so the y-component is simply vy = -5.39 m/s.
Step 2: Calculate the y-component when the velocity vector is at a downward angle of 45 degrees.
Since the angle between the velocity vector and the y-axis is 45 degrees, the y-component can be calculated using the formula:
y-component = vy * sin(angle)
= -5.39 m/s * sin(45)
= -5.39 m/s * 0.707 (approximating sin(45) as 0.707)
= - 3.81 m/s (approximated to two decimal places)
Therefore, the y-component of the velocity will be approximately -3.81 m/s when Tom's velocity vector is pointing down at an angle of 45 degrees.