+8262 When you go camping, bears always have a good sense of smell and they can eat your food. If you throw a sack with food on top of a branch that is 10 feet above sea level and the velocity that you throw it is 28 feet per second, will it arrive over the branch so the bear can not reach the food? If it does, how much time will it take for it to get on top of the branch? Explain.
Please help.
+4473 
We must take into account the impact of gravity.
By plugging in v(initial) as $$\frac{28 ft}{sec}$$, d as 10 ft, and acceleration(a) as $$\frac{-9.8 m}{sec^{2}} = \frac{-32.2 ft}{sec^{2}}$$ we obtain:
$$10 ft = \frac{28ft}{sec}t + \frac{1}{2}\frac{-32.2 ft}{sec^{2}}t^{2}$$
We could rearrange this equation and use the quadratic equation to solve for time now, t; however, let's try some other equation and find a time, t, to plug back into the first equation ():
$$v_{f} = v_{i} + at$$
Let's assume the v(final) will be 0 since the sack has to stop moving. Then:
$$0 = \frac{28 ft}{sec} + \frac{-32.2 ft}{sec^{2}}t$$
$$\frac{-28 ft}{sec} = \frac{-32.2ft}{sec^{2}}t$$
$$t = \frac{\frac{-28 ft}{sec}}_{\frac{-32.2 ft}{sec^{2}}$$
t = 0.8695652173913043 seconds
Notice that if we plug in this time, t, in the first equation --> $$10 ft = \frac{28ft}{sec}t + \frac{1}{2}\frac{-32.2 ft}{sec^{2}}t^{2}$$, we would not get a distance of 10 ft, but 12.1739130434782608696 feet.
This means that it would go over the branch(?).
The time it would take to get on the branch would be the time, t, we would get by solving $$10 ft = \frac{28ft}{sec}t + \frac{1}{2}\frac{-32.2 ft}{sec^{2}}t^{2}$$ by using the formula below with a = -16.1, b = 28 & c = -10 -->
$$\begin{array}{*{20}c} {t = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{}}} \\ \end{array}$$ (Thanks Phill for this LaTex Formula!)
t = 1.237023589, 0.5021068458.....
t = 0.5021068458 seconds makes the most sense because at 12.17 feet, it would have taken 0.87 seconds. For 10 feet, it would take less than <0.87 seconds, so a time greater than 1 second would not make sense.
**************************************************************************************
So, the sack it will not arrive on top of the branch, but will instead go over it(?). However, it will pass the branch at around [[t = 0.5021068458 seconds]].
I am not sure if this is how the question is supposed to be solved...but I tried this method and I think I may have confused myself more!
+4473
We must take into account the impact of gravity.
By plugging in v(initial) as $$\frac{28 ft}{sec}$$, d as 10 ft, and acceleration(a) as $$\frac{-9.8 m}{sec^{2}} = \frac{-32.2 ft}{sec^{2}}$$ we obtain:
$$10 ft = \frac{28ft}{sec}t + \frac{1}{2}\frac{-32.2 ft}{sec^{2}}t^{2}$$
We could rearrange this equation and use the quadratic equation to solve for time now, t; however, let's try some other equation and find a time, t, to plug back into the first equation ():
$$v_{f} = v_{i} + at$$
Let's assume the v(final) will be 0 since the sack has to stop moving. Then:
$$0 = \frac{28 ft}{sec} + \frac{-32.2 ft}{sec^{2}}t$$
$$\frac{-28 ft}{sec} = \frac{-32.2ft}{sec^{2}}t$$
$$t = \frac{\frac{-28 ft}{sec}}_{\frac{-32.2 ft}{sec^{2}}$$
t = 0.8695652173913043 seconds
Notice that if we plug in this time, t, in the first equation --> $$10 ft = \frac{28ft}{sec}t + \frac{1}{2}\frac{-32.2 ft}{sec^{2}}t^{2}$$, we would not get a distance of 10 ft, but 12.1739130434782608696 feet.
This means that it would go over the branch(?).
The time it would take to get on the branch would be the time, t, we would get by solving $$10 ft = \frac{28ft}{sec}t + \frac{1}{2}\frac{-32.2 ft}{sec^{2}}t^{2}$$ by using the formula below with a = -16.1, b = 28 & c = -10 -->
$$\begin{array}{*{20}c} {t = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{}}} \\ \end{array}$$ (Thanks Phill for this LaTex Formula!)
t = 1.237023589, 0.5021068458.....
t = 0.5021068458 seconds makes the most sense because at 12.17 feet, it would have taken 0.87 seconds. For 10 feet, it would take less than <0.87 seconds, so a time greater than 1 second would not make sense.
**************************************************************************************
So, the sack it will not arrive on top of the branch, but will instead go over it(?). However, it will pass the branch at around [[t = 0.5021068458 seconds]].
I am not sure if this is how the question is supposed to be solved...but I tried this method and I think I may have confused myself more!
+8262 AzizHusain, I re-edited my post because I missed out a few words.
I wanted to know that if you throw it over the branch, can you tie it to the branch so the bear can't eat the food?
+4473 Yes, it will arrive over the branch so the bear cannot reach the food. You can tie it as well.
I am imagining something like this:
