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October 27th, 2019, 07:50 PM   #1
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Question How can I find the distance covered by a stuntman jumping over an incline?

The problem is as follows:
A stuntmant in a circus act rides on a motorbike as it is shown in the figure from below. The bike which accelerates $10 \frac{m}{s^{2}}$ goes over an incline forming a $53^{\circ}$ angle with the horizontal. Find from what distance (given on meters) from point A to the ground the motorist makes an impact. (You may use $g=10\frac{m}{s^{2}}$ and assume a $3-4-5$ triangle with opposing angles $37-53-90$ respectively).


The alternatives given on my book are as follows:

$\begin{array}{ll}
1.&25\\
2.&75\\
3.&25\sqrt{5}\\
4.&25\sqrt{10}\\
5.&10\sqrt{13}\\
\end{array}$

Supposedly according to my book the answer is the option number $4$. However I am unable to get to that answer.

This problem shouldn't be complicated but each time I go after the steps I attempted I dont't get any close to the alledged answer.

What I did was the following:

The height from the top of the ramp to the ground is: (For brevity purposes I'm omitting the units)

Given:

$5k=30$

$k=6$

Therefore:

$4k+1=4\times 6 + 1 = 25$

Since what they are asking is the position where the stuntman will impact to the ground the equation that I will use is:

$y(t)= y_{o}+v_{o}\sin\omega t -\frac{1}{2}gt^2$

This becomes into:

$y(t)= 25 +v_{o}\sin\omega t-\frac{1}{2}gt^2$

However I need the initial velocity from which the motorist will exit the ramp right at the start of the jump.
What I did for this part was this:

$v_{f}^{2}=v_{o}^{2}+2a\Delta x$

Therefore replacing the information given: ($a=10$ and $v_{o}=5$ and $x=30$)

This will become into:

$v_{f}^{2}=\left(5\right)^{2}+2\left(10\times 30\right) = 625$

$v_{f}=25$

So with that velocity I can plug in the previous equation and from that I can find the elapsed time that it took the whole jump from that given height to reach $0$ that is the ground.

By replacing that it will become into:

$y(t)= 25 +25\sin 53^{\circ}t-\frac{1}{2}\left(10\right)t^2$

$y(t)= 25 +25\left(\frac{4}{5}\right)t-\frac{1}{2}\left(10\right)t^2$

$y(t)= 25 +20t-5t^2$

So when $y(t)=0$

$5t^2-20t-25=0$

$t^2-4t-5=0$

From this information it can be obtained that:

$t=\frac{4\pm\sqrt{36}}{2}=2\pm 3$

$t=5$

Now all is left to do is to plug this time in the equation for the horizontal component ($x-axis$)

$x=v_{o}\cos\omega t$

$x=25\times \left(\frac{3}{5}\right) \times 5= 75$

Therefore the distance would be $75$.

But am I wrong with the way how I used the logic in the problem or could it be that there exist another thing to do?.

Can somebody help me with this problem?
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October 27th, 2019, 08:12 PM   #2
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$v_f^2 = v_0^2 + 2a(\Delta s)$

at the top of the incline ...

$v_f = \sqrt{5^2 + 2(10)(30)} = 25 \, m/s$

$v_{y0} = v_f \cdot \sin{\theta} = 25 \cdot \dfrac{4}{5}=20 \, m/s$


$\Delta y = v_{y0} \cdot t - \dfrac{1}{2}gt^2$

$-25 = 20t - 5t^2 \implies t^2 - 4t - 5 = 0 \implies (t-5)(t+1)=0 \implies t = 5$


$\Delta x = v_f \cdot \cos{\theta} \cdot t = 25 \cdot \dfrac{3}{5} \cdot 5 = 75 \, m$

I agree with your solution ... #2
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October 28th, 2019, 10:14 PM   #3
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Quote:
Originally Posted by skeeter View Post
$v_f^2 = v_0^2 + 2a(\Delta s)$

at the top of the incline ...

$v_f = \sqrt{5^2 + 2(10)(30)} = 25 \, m/s$

$v_{y0} = v_f \cdot \sin{\theta} = 25 \cdot \dfrac{4}{5}=20 \, m/s$


$\Delta y = v_{y0} \cdot t - \dfrac{1}{2}gt^2$

$-25 = 20t - 5t^2 \implies t^2 - 4t - 5 = 0 \implies (t-5)(t+1)=0 \implies t = 5$


$\Delta x = v_f \cdot \cos{\theta} \cdot t = 25 \cdot \dfrac{3}{5} \cdot 5 = 75 \, m$

I agree with your solution ... #2
Supposedly it seems that what is being asked is the straight distance from $A$ to the ground, in other words the hypotenuse of the right triangle.

Therefore $\sqrt{75^{2}+25^{2}}=25\sqrt{10}$

which would check with the fourth alternative.

Now, looking at a real life scenario.

Have you ever attended to such acts?. I mean those like Evel Knievel made famous?.

Personally I never seen them in person, but I had watched them on tv and always fascinated me. However this problem doesn't really take into account that what people does matter is how far or how tall the stuntman will reach, as when you look at the faces of the spectators they're looking up and down, but I believe never they do look diagonally to see such "trajectory". I think it is some silly to ask this as well to my opinion is not very realistic.
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