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).

Sketch of the problem

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)





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

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:

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


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 $y(t)=0$



From this information it can be obtained that:

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


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?


They ask you the distance "from point A". You calculated the horizontal distance as 75 meters, the vertical distance as 25 meters, so all you need to do is use Pythagoras' theorem $$d=\sqrt{25^2+75^2}=25\sqrt{1+3^2}=25\sqrt{10}$$

  • $\begingroup$ It seems I overlooked that fact. Now that I think about it. I didn't know what was meant by the distance from that point. Thanks for that!. $\endgroup$ – Chris Steinbeck Bell Oct 28 '19 at 4:11
  • $\begingroup$ Happens to me as well sometimes. Don't forget to mark the answer as accepted $\endgroup$ – Andrei Oct 28 '19 at 4:43

I believe you’re correct

  1. The height when leaving the ramp is:

$y_0=1+30\cdot sin(53^0)\approx 25$

  1. The distance tavelled horizontally is:

$x=v_0\cdot cos(\alpha)\cdot t$

  1. Vertically:

$y=y_0 +v_0\cdot sin(\alpha)\cdot t-\frac{g\cdot t^2}{2}$

  1. Replace t from step 2 and obtain a quadratic in x; solve for y=0
  • $\begingroup$ So by solving for $x$ I'm obtaining automatically the range?. Is this what you meant?. How should I understand the negative value from that quadratic?. Or does that value become ignored?. $\endgroup$ – Chris Steinbeck Bell Oct 28 '19 at 4:10
  • $\begingroup$ I thought so, however it seems not the horizontal distance but the actual distance between launch and landing was required. As for the negative root, the quadratic gives the roots and not the times, so obviously you will not consider the negative time value. $\endgroup$ – WindSoul Oct 28 '19 at 5:22
  • $\begingroup$ Noted. It was an interesting approach as it gives the distance directly, however they are asking the diagonal. In other words the hypotenuse of the right triangle. $\endgroup$ – Chris Steinbeck Bell Nov 1 '19 at 23:42

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