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

Jun 2017
337
6
Lima, Peru
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 $2\,\frac{m}{s}$. 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 misinterpreting 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?!! Is there an error in my understanding of the problem or could it be that the alternatives aren't given right?
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
$\dfrac{\Delta y}{\Delta x} = -\dfrac{1}{2} \implies \Delta y = -\dfrac{\Delta x}{2}$

$\Delta y = 1.2 t - 5t^2$, $\Delta x = 1.6t$

$1.2t - 5t^2 = -0.8t \implies 2t - 5t^2 = 0 \implies t(2-5t) = 0 \implies t = 0.4 \text{ sec}$

check that value for $v_0$ ... 2 m/s is only about 4.5 mph. I'm old, but I can run faster than that. If $v_0$ is 20 m/s, then the cycle will stay airborne for 4 sec.
 
Jun 2017
337
6
Lima, Peru
$\dfrac{\Delta y}{\Delta x} = -\dfrac{1}{2} \implies \Delta y = -\dfrac{\Delta x}{2}$

$\Delta y = 1.2 t - 5t^2$, $\Delta x = 1.6t$

$1.2t - 5t^2 = -0.8t \implies 2t - 5t^2 = 0 \implies t(2-5t) = 0 \implies t = 0.4 \text{ sec}$

check that value for $v_0$ ... 2 m/s is only about 4.5 mph. I'm old, but I can run faster than that. If $v_0$ is 20 m/s, then the cycle will stay airborne for 4 sec.
I'm sorry for delaying in this response, but I had to consult with the original source and you're right. There was an error in this problem. The speed of the motorist was $20\,\frac{m}{s}$, and the flying time is $4\,s$ as the answer.

The thing which I'm still confused about is why the height which I assigned in the upper sketch isn't right?. Why can't be assumed that the vertical height (from where the motorist leavers the ramp) and I'm referring to $y_{o}=2t\sin 37^{\circ}$ ? Can you help me with that part?
 

skeeter

Math Team
Jul 2011
3,266
1,766
Texas
I'm sorry for delaying in this response, but I had to consult with the original source and you're right. There was an error in this problem. The speed of the motorist was $20\,\frac{m}{s}$, and the flying time is $4\,s$ as the answer.

The thing which I'm still confused about is why the height which I assigned in the upper sketch isn't right?. Why can't be assumed that the vertical height (from where the motorist leavers the ramp) and I'm referring to $y_{o}=2t\sin 37^{\circ}$ ? Can you help me with that part?
You have the equation $y = {\color{red}2t\sin(37)} + v_0\sin(37)t - {\large\frac12}gt^2$

The first term (highlighted in red) is not affected by acceleration due to gravity, whereas the last two terms together are affected by gravity.
Let's say it takes $t = 3$ seconds for the cycle to climb the ramp, does the gravity term, $-5t^2 = -45$ make sense if the cycle hasn't left the ramp?

All the basic kinematics equations only work if acceleration is uniform, i.e. the same.