Question

1. A cat jumps off of the table with a velocity v0 at an angle θ. The table is h meters tall. Write the
velocity vector as a function of time. v(t)=?

2. What is the angle with respect to the ground that the cat makes when it hits the ground?

I got number one I will write it out in a second. I don't exactly understand the 2nd question. (Our teacher makes us solve problems with only variables sometimes )

Answer 1

What I got for number 1
\[\frac{ -v _{0}\sin \Theta \pm \sqrt{v _{0}\sin \Theta ^{2} + 19.6h}}{ -9.8 }\]

Answer 2

That's the time in which the thing would hit the ground (with the minus sign), not the velocity vector. The velocity vector as a function of time is
\[ \vec{v}(t) = <v_0 \cos(\theta), v_0 \sin(\theta) - gt >\]

Answer 3

I don't really get the velocity vector

Answer 4

@Jemurray3 why is \[V _{0}\sin(\theta) -gt\] in the vector

Answer 5

The x-component of the velocity is constant, so it's just v cos( theta ). The y component starts off at v sin( theta ) and decreases from there. Velocity = Initial velocity + acceleration * time, right?

Answer 6

Ohhhh okay @Jemurray3 I get that part! Thank you . Could you help with number 2? I don't understand it at all

Answer 7

So would time also be with VCos(theta) ? @jemurray3

Answer 8

As I said, the x-component of the velocity is constant, so no, it would not. As for number two, when the cat hits the ground:

Answer 9

ok ok I gotcha

Answer 10

It wants to know that angle. So first, find the time it takes to hit the ground (you already did most of the work for that part in your first try ). Then, plug that in to find the y component of the velocity at that time. You'll get a negative number. Then, take
\[ \theta = \tan^{-1} \left( \frac{|v_y|}{|v_x|}\right) \]

Answer 11

Wait so my first post was time? Was it not?

Answer 12

@Jemurray3

Answer 13

@jemurray3 I got the y component to be \[-V _{0}\sin \Theta -\sqrt{V _{0}\sin \Theta ^{2} + 19.6h}\]

Answer 14

Disregard the - VSin(theta) at the beginning