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?