How to find the flying time of a motorist going over an incline with the shape of a triangle?

Question

The problem is as follows:

A stuntman in a circus act rides on his motorbike over a ramp as seen in the figure from below. The motorbike has a motion with no acceleration and its speed is of $20\,\frac{m}{s}$. How long will he stay on the air?. Assume that the acceleration due gravity is $10\,\frac{m}{s}$ and $\tan\alpha=\frac{1}{2}$.

The alternatives given are as follows:

$\begin{array}{ll} 1.&2\,s\\ 2.&3\,s\\ 3.&4\,s\\ 4.&5\,s\\ 5.&6\,s\\ \end{array}$

What I attempted to so is summarized in the sketch from below:

The motorbike ascends to the top of the ramp and from there jumps. Since the tangent of the opposite angle in the incline is given, then it can be established the horizontal distance.

From this I obtained the equation for the flying time in the vertical component as follows:

$y=2t\sin 37^{\circ}+2t\sin 37^{\circ}-5t^2$

$y=4t\sin 37^{\circ}-5t^2$

$y=4t\frac{3}{5}-5t^2$

$0=\frac{12}{5}t-5t^2$

$0=12t-25t^2$

$t(12-25t)=0$

$t=\frac{25}{12}$

However this doesn't seem along the answers. What exactly could I be missinterpreting here?.

If I do establish the equation for the horizontal component it ends as follows:

$x=2\cos 37^{\circ}t$

Then:

$\frac{2t\sin37^{\circ}}{\tan\alpha}=2\cos 37^{\circ}t$

However this produces an inconsistency.

But if I were to use the equation in terms of $x$ and $y$ this becomes into:

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

$y\left(\frac{x}{v_o\cos\omega}\right)=y_{o}+v_o\sin\omega\frac{x}{v_o\cos\omega }-\frac{1}{2}g\left(\frac{x}{v_o\cos\omega}\right)^2$

$y=y_{o}+x\tan\omega-\frac{1}{2}g\left(\frac{x}{v_o\cos\omega}\right)^2$

Given the conditions of the problem: $y_{o}=2\sin37^{\circ}t=\frac{6t}{5}$

$x=\frac{2t\sin37^{\circ}}{\tan\alpha}=\frac{12t}{5}$

$y=\frac{6t}{5}+\frac{12t}{5}\tan 37^{\circ}-5\left(\frac{\frac{12t}{5}}{2\cos37^{\circ}}\right)^2$

$y=\frac{6t}{5}+\frac{12t}{5}\left(\frac{3}{4}\right)-5\left(\frac{\frac{12t}{5}}{\frac{8}{5}}\right)^2$

$y=\frac{6t}{5}+\frac{9t}{5}-\frac{45t^2}{4}$

$y=\frac{15t}{5}-\frac{45t^2}{4}$

$y=3t-\frac{45t^2}{4}$

Finally: $y=0$

$0=3t-\frac{45t^2}{4}$

$0=12t-45t^2$

$0=4t-15t^2$

$0=t(4-15t)$

Therefore:

$t=\frac{15}{4}$

Therefore where did I made an error can somebody help me here?!!

Answer 1

There is something wrong with the statement of this problem. The diagram is fairly misleading, because the far side of the ramp looks much steeper than a slope with $\alpha=\tan^{-1}0.5\approx27^\circ$.

But referring to the text and Fig 1 below it, the cyclist is about to leave the ramp with speed 2. Given the angle $37^\circ$ that is an initial upward speed of 1.2 and a horizontal speed of 1.6. The horizontal speed will not change during the motion.

[The following day, the initial speed was changed, see the end of this answer. At the time of writing the question had been partially edited to reflect the change.]

Take rectangular coordinates with origin at the top of the ramp, $x$-axis horizontal and $y$-axis vertically upwards. Then we have $$x(t)=1.6t,y(t)=1.2t-5t^2\quad(*)$$

It is not clear from Fig 1 or the problem text how tall the ramp is. Its surface is described by the line $y=-x\tan\alpha=-0.5x$. So his trajectory will hit the ramp at time $t$ where $$1.2t-5t^2=y(t)=-0.5x(t)=-0.8t$$ and hence $t=0$ or $t=0.4$. The first value is just his takeoff point at the top of the ramp. The second value is at a distance 0.64 below the top of the ramp.

So the apparent answer to the question is $t=0.4$ which is not one of the multiple choice answers.

If the ramp was less than 0.64 high, then he would presumably hit the ground the far side of the ramp. But that would be at an earlier point of the trajectory (imagine the slope of the ramp going underground - he hits the ground before reaching the ramp again) and hence at an earlier time.

Another possibility is that we are meant to include his time driving up the ramp, but that does not fit well with the wording "stay in the air".

So I conclude that the question/solution are badly drafted. Note also the words "the motorbike has no acceleration". What does that mean? Does it mean no gravity (in which case he never comes down). Or that he freewheeled up the ramp?

Is Fig 2 part of the official solution? If so, I do not understand the $y=2t\sin37^\circ+v_0t\sin37^\circ-5t^2$ because $v_0=2$, so that term has (wrongly) been included twice. But that error does not explain everything. If we changed the initial speed to 4, it the time of impact would still only be $t=0.84$.

All in all a complete mess!

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Added later.

So it turns out that the initial speed was 20, not 2. So $(*)$ becomes $$x(t)=16t,y(t)=12t-5t^2,\text{ or }5t^2=20$$ and hence $t=2s$ when he hits the downramp .

Anything to learn? Yes! When applying math to the real world one needs to check plausibility. 2m/s is a pathetic speed for anyone on a bike etc - it can be easily matched by a fast runner. 20m/s is much more plausible! I should have noticed that!

Answer 2

You have a diagram with the height of the ramp labeled as $2t\sin 37^\circ.$ That's correct if $t$ is the amount of time the motorcyclist spends climbing the ramp from the bottom to the top. However, since the amount of time spent climbing the ramp is not part of what the question asked for, this is a fairly useless fact. Let's forget that this formula was ever written.

We also have to throw away any calculations where $2t\sin 37^\circ$ got inserted, and any further results that depend on those calculations. That's pretty much everything you wrote, unfortunately.

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An alternative method:

Recalling that $37^\circ$ is approximately the angle of a $3$-$4$-$5$ right triangle and that physics teachers love to assume the approximation is exact, we have $\sin 37^\circ = \frac35$ and $\cos 37^\circ = \frac45.$ We therefore know that the initial components of the motorcycle's velocity are

\begin{align} v_{x0} &= 2 \cos 37^\circ = 2\left(\frac45\right) = \frac 85, \\ v_{y0} &= 2 \sin 37^\circ = 2\left(\frac35\right) = \frac 65. \\ \end{align}

Now imagine a hypothetical insect running down the right side of the ramp at a constant horizontal velocity of $\frac85,$ starting from the top at the instant the motorcycle goes airborne. The insect therefore will be directly underneath the motorcycle for the entire time until the motorcycle lands on the ramp again, "squashing" the insect. (Remember, it's only a hypothetical insect; no real insects were harmed in the making of this answer.)

Because the right side of the ramp is at an angle whose tangent is $\frac12,$ the insect's vertical velocity is half the magnitude of its horizontal velocity, so its vertical velocity is $-\frac45.$

Now let's make a change in frame of reference. The new frame is moving at a vertical velocity $-\frac45$ with respect to the frame of rest of the ramp. The new frame is still an inertial frame because it differs from the original inertial frame only by a constant linear velocity, but the in the new frame the bug is running along a horizontal path. We therefore are simply looking for the flight time of the motorcycle for a simple parabolic trajectory that starts and ends at the same height.

In the new frame, the motorcyclist's initial vertical velocity is

$$ v'_{x0} = v_{x0} + \frac45 = \frac 65 + \frac 45 = 2. $$

The total flight time is

$$ \frac{2v'_{x0}}{g} = \frac{2\times2}{10} = 0.4. $$

This is not one of the answers. Working backward from the smallest answer, $2$ seconds, we see that the motorcyclist's vertical velocity needs to be at least $10$ m/s greater than the bug's in order to achieve that flight time. In other words we need to speed everyone up by a factor of at least $5$ to get any of the times in the answer choices.

Is a decimal point missing somewhere? Is the answer "$4$s" actually $.4$s? Or perhaps the $2$ m/s was supposed to be $20$ m/s? (The latter would make more sense for a motorcycle stunt. Two meters per second is barely faster than walking.)

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Now, I happen to have a tendency to change reference frames in order to convert problems I don't like into problems that I like, but I suspect the "book" solution doesn't do this. What you might be expected to do is to use the formula you already know in order to describe the motorcycle's trajectory:

$$y = x\tan\phi - \frac12 g\left(\frac{x}{v_o\cos\phi}\right)^2. \tag1$$

You can use this formula directly if you define your coordinates so that the top of the ramp is at $(0,0).$ Then the formula for the right side of the ramp is

$$y = -\frac12 x. \tag2$$

Obviously the line described by $(2)$ intersects the parabola described by $(1)$ at $(0,0).$ You can find the other intersection using algebra. You then have to determine the time at that intersection. The simplest way is to divide the horizontal distance (which happens to be the $x$ coordinate of the intersection, since the jump starts at $x=0$) by the horizontal speed.

Note how different this is from the solution in the other reference frame. In the other solution I never considered what the path of the motorcycle is or what horizontal distance it travels. I did not even use the horizontal velocity; I used the initial vertical velocity, which worked nicely since in that frame I merely needed to know the time taking to return to the same height. I did not even need to say what $y$ coordinate I started from.