- #1

thegreengineer

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Well people, this is the given problem:

Your friend is trying to impress his girlfriend jumping a river that is 10.5 m long (this is the

[itex]R=\frac{(v_{0})^{2}(\textbf{sin}(2θ))}{g}[/itex]

[itex]y_{max}=\frac{(v_{0})^{2}(\textbf{sin}^{2}(θ))}{2g}[/itex]

[itex]t=\frac{2(v_{0})(\textbf{sin}(θ))}{g}[/itex]

In which R is the range, ymax is the maximum height, and t is the time that it took in the air.

Ok, so my main problem is to determine whether this guy reaches the cliff or not. According to what I've got in the data I was able to determine the range which resulted in:

[itex]R=\frac{(12\frac{m}{s})^{2}(\textbf{sin}(2(30°)))}{9.81\frac{m}{s^{2}}}=12.73 m[/itex]

Now my main problem is that even though I found that the range is larger than what the river is 10.5 m long; we can assume that this guy will make it to the other side. Now to prove that is to find that he can reach it in vertical terms (I mean, that he won't be falling short and he will reach the other side without crashing)I need to find if he makes it or not because I don't feel secure saying it just finding that the range was larger than the river's length.

Your friend is trying to impress his girlfriend jumping a river that is 10.5 m long (this is the

**horizontal range**). Your friend wants you to recommend him how to improve the jump as he tells you that the jumping point (the cliff from which your friend wants to jump) is 3 m above the river as it has an inclination angle of 30° (with respect to x axis). The other point to reach (i.e. the other cliff) is 4 m above the river. If the motorcycle's max velocity that can run is 12 m/s, would you recommend him to jump or not? Justify your answer.Ok people so this is parabolic motion, even though we already know the equations of motion for constant acceleration, I'm only going to write the ones concerning with parabolic motion which as I saw in school are:[itex]R=\frac{(v_{0})^{2}(\textbf{sin}(2θ))}{g}[/itex]

[itex]y_{max}=\frac{(v_{0})^{2}(\textbf{sin}^{2}(θ))}{2g}[/itex]

[itex]t=\frac{2(v_{0})(\textbf{sin}(θ))}{g}[/itex]

In which R is the range, ymax is the maximum height, and t is the time that it took in the air.

Ok, so my main problem is to determine whether this guy reaches the cliff or not. According to what I've got in the data I was able to determine the range which resulted in:

[itex]R=\frac{(12\frac{m}{s})^{2}(\textbf{sin}(2(30°)))}{9.81\frac{m}{s^{2}}}=12.73 m[/itex]

Now my main problem is that even though I found that the range is larger than what the river is 10.5 m long; we can assume that this guy will make it to the other side. Now to prove that is to find that he can reach it in vertical terms (I mean, that he won't be falling short and he will reach the other side without crashing)I need to find if he makes it or not because I don't feel secure saying it just finding that the range was larger than the river's length.

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