Volleyball Impulse question

[sheepcountme]

Homework Statement

A ball is traveling horizontally over a volleyball net when a player "spikes" it, driving it straight down to the ground. The ball's mass is 0.20 kg, its speed before being hit is 6.4 m/s and its speed immediately after the spike is 21 m/s. What is the magnitude of the impulse from the spike?

Homework Equations

J=mv(final)-mv(initial)

The Attempt at a Solution

J=.20(0,-21)-.20(6.4,0)
J=(0,-4.2)-(1.28,0)
J=(1.28,-4.2)

I don't understand where I went wrong, but the book tells me that this is incorrect.

 

[cepheid]

Well, assuming your calculations are correct, that's the impulse in vector form. The book wants the magnitude of the impulse.

 

[sheepcountme]

Ah, perhaps you can tell me where I went wrong initially...

My first attempt to solve the problem was:
.20(21)-.20(6.4)=4.2-1.28=2.92
2.92 was incorrect, and then I thought perhaps it should be negative since the direction is down, but -2.92 was incorrect as well, so that is when I tried the vector form..which, to my dismay, was also incorrect.

 

[cepheid]

Sorry for not getting back to you earlier.

Ah, perhaps you can tell me where I went wrong initially...

My first attempt to solve the problem was:
.20(21)-.20(6.4)=4.2-1.28=2.92

Well, just algebraically adding together the x and y components of a vector doesn't make much sense, does it?

2.92 was incorrect, and then I thought perhaps it should be negative since the direction is down, but -2.92 was incorrect as well, so that is when I tried the vector form..which, to my dismay, was also incorrect.

It's not that your answer for the impulse vector was necessarily wrong. It's just that the question wasn't asking for the vector. It was asking for the magnitude of that vector. How do you compute the magnitude of a vector? Hint: If you draw a picture of the x-component and y-component of a vector adding up "tip to tail" to form the resultant vector, what you have drawn is a right-angled triangle. The magnitude of the vector is its length, which is the length of the hypotenuse of the triangle in your picture. I think the calculation in your original post is mostly right, except...

The Attempt at a Solution

J=.20(0,-21)-.20(6.4,0)
J=(0,-4.2)-(1.28,0)
J=(1.28,-4.2)

How do you get to this last line? For the x-component, 0 - 1.28 is not equal to 1.28. Also, you know that the x-component of the impulse should be negative, since the x-component of the momentum is initially positive, and ends up being zero.