Which choice do I choose from the quadratic?

[Rijad Hadzic]

Homework Statement

A circus cat has been trained to leap off a 12 m high platform and land on a pillow. The cat leaps off at $v_0 = 3.5 m/s$ at an angle of 25 degrees. Where should the trainer place the pillow so that the cat lands safely? What is the cats velocity as she lands on the pillow?

Homework Equations

$\Delta x = (1/2)(V_{0x} + V_x)t$

$\Delta x = V_{0x}t + (1/2)a_x t^2$

The Attempt at a Solution

So I calculated $V_{0y} = 3.5sin(25) = 1.479163916 m/s$

and

$V_{0x} = V_x = 3.5cos(25) = 3.172077255 m/s$

So there are two times here,

what I will call $t_1$ = time it takes from start of movement to very top, and then back down, so a $\Delta y$ displacement of 0.

$t_2$ = From the above, the second part, starting when the y displacement hits 0, to when the cat lands on the pillow.

I use equation $V_{0x} + a_x t$ to find $t_1$

Since I know the inital y velocity = 1.479163916, I know that at the point I'm talking about, after it goes up and comes back down to a $\Delta y$ of 0, the velocity is going to be the same magnitude but opposite direction, so final y velocity = -1.479163916

solving for t I get

$t_1 = (-1.479163916 - 1.479163916 ) / -9.8 = .3018701869$

Does this make sense so far?

Now I have to get $t_2$ so I use formula: $\Delta y = V_{0y}t + (a_y/2)t^2$

with $V_{0x} = -1.479163916 m/s$ since we are starting AFTER $t_1$, $a_x/2 = -4.9 m/s^2$ and $\Delta y = 12 m$

Now I want to use the quadratic equation to get my t here, but under the square root I get a negative value. my quadratic looks like:

$-V_{0y} {+-} \sqrt {V_{0y}^2 -4(a_y/2)(-\Delta y)}$ all divded by $a_y$

is what my quadratic looks like. Plugging the values I listed above for the variables, I get a - inside my sqroot which makes no sense to me. I realize that this IS the way to go about this problem though, I just don't understand what I did wrong from here?

 

[kuruman]

I don't know what you used for a_x. There is no acceleration in the x-direction. I suggest that you write an expression giving the position of the cat above the pillow at any time t and then find at what time the cat is on the pillow. This is more direct than doing it in two segments which leaves you open for additional errors.

 

[Rijad Hadzic]

I don't know what you used for a_x. There is no acceleration in the x-direction. I suggest that you write an expression giving the position of the cat above the pillow at any time t and then find at what time the cat is on the pillow. This is more direct than doing it in two segments which leaves you open for additional errors.

Sorry I meant to say I used formula
$\Delta y = V_{0y}t + (a_y/2)t^2$ and ay was = -9.8 m/s^2... for time 2.

I didn't use any x velocity in my calculations so far at all, that was just a typo..

Although your way does seem more direct, I didn't think of that way myself.. Is it still possible to get the answer the way I'm doing it?

 

[kuruman]

Is it still possible to get the answer the way I'm doing

Yes it is, but the more steps you introduce, the more likely you are to make a mistake. Just solve the quadratic you quoted in your previous post and you are done.

 

[Ray Vickson]

Homework Statement

A circus cat has been trained to leap off a 12 m high platform and land on a pillow. The cat leaps off at $v_0 = 3.5 m/s$ at an angle of 25 degrees. Where should the trainer place the pillow so that the cat lands safely? What is the cats velocity as she lands on the pillow?

Homework Equations

$\Delta x = (1/2)(V_{0x} + V_x)t$

$\Delta x = V_{0x}t + (1/2)a_x t^2$

The Attempt at a Solution

So I calculated $V_{0y} = 3.5sin(25) = 1.479163916 m/s$

and

$V_{0x} = V_x = 3.5cos(25) = 3.172077255 m/s$

So there are two times here,

what I will call $t_1$ = time it takes from start of movement to very top, and then back down, so a $\Delta y$ displacement of 0.

$t_2$ = From the above, the second part, starting when the y displacement hits 0, to when the cat lands on the pillow.

I use equation $V_{0x} + a_x t$ to find $t_1$

Since I know the inital y velocity = 1.479163916, I know that at the point I'm talking about, after it goes up and comes back down to a $\Delta y$ of 0, the velocity is going to be the same magnitude but opposite direction, so final y velocity = -1.479163916

solving for t I get

$t_1 = (-1.479163916 - 1.479163916 ) / -9.8 = .3018701869$

Does this make sense so far?

Now I have to get $t_2$ so I use formula: $\Delta y = V_{0y}t + (a_y/2)t^2$

with $V_{0x} = -1.479163916 m/s$ since we are starting AFTER $t_1$, $a_x/2 = -4.9 m/s^2$ and $\Delta y = 12 m$

Now I want to use the quadratic equation to get my t here, but under the square root I get a negative value. my quadratic looks like:

$-V_{0y} {+-} \sqrt {V_{0y}^2 -4(a_y/2)(-\Delta y)}$ all divded by $a_y$

is what my quadratic looks like. Plugging the values I listed above for the variables, I get a - inside my sqroot which makes no sense to me. I realize that this IS the way to go about this problem though, I just don't understand what I did wrong from here?

I think you need $\Delta y = -12$ in the equation for $t_2$.

Here is how I would do it if I adopted your approach:
To get $t_1$, solve
$$12 =12 + V_{0y} t - \frac{1}{2} g t^2\hspace{1cm}(1)$$
and to get $t_2$ solve
$$0 = 12 - V_{0y} t - \frac{1}{2} g t^2 \hspace{1cm} (2) $$
However, as 'kurman' suggests, you can get $t_0 = t_1 + t_2$ directly by solving
$$0 = 12 + V_{0y} t - \frac{1}{2} g t^2 \hspace{1cm} (3)$$
Note that the only difference between (2) and (3) is the sign of the $V_{0y}t$ term on the right.

The positive root of (2) is the one you want; you might want to think about the meaning (if any!) of the negative root. Ditto for eq. (3).

 

[Rijad Hadzic]

I think you need $\Delta y = -12$ in the equation for $t_2$.

Here is how I would do it if I adopted your approach:
To get $t_1$, solve
$$12 =12 + V_{0y} t - \frac{1}{2} g t^2\hspace{1cm}(1)$$
and to get $t_2$ solve
$$0 = 12 - V_{0y} t - \frac{1}{2} g t^2 \hspace{1cm} (2) $$
However, as 'kurman' suggests, you can get $t_0 = t_1 + t_2$ directly by solving
$$0 = 12 + V_{0y} t - \frac{1}{2} g t^2 \hspace{1cm} (3)$$
Note that the only difference between (2) and (3) is the sign of the $V_{0y}t$ term on the right.

The positive root of (2) is the one you want; you might want to think about the meaning (if any!) of the negative root. Ditto for eq. (3).

Thanks for the response. I think I'm starting to understand now.

 

[Ray Vickson]

Thanks for the response. I think I'm starting to understand now.

While I have your attention: first, thanks for using LaTeX to type your solution. However, can I suggest some improvements?
(1) Never use "sin" or "cos", which produce ugly results that are hard to read; always use "\sin" or "\cos", as these have been designed to produce pleasing output: for instance, $sin a$ vs. $\sin a$. The same goes for most of the common short functions: all the trig and inverse trig functions, the functions 'exp', 'log', 'ln', the hyperbolic functions (but not their inverses!), the instructions 'lim' (for limit), 'inf' (for infimum), 'sup' (supremum), 'max, ' min' and some others like 'gcd' (greatest common divisor), etc.
(2) If you are not using radians to measure angles, be sure to Include the units of angular measurement, as in $\sin(25^o)$ or $\sin(25^{\circ}).$