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 $\sin(25^{\circ}).$