How can I find the flying time for a projectile fired from some elevation and hitting an incline at a given angle?

Question

The problem is as follows:

In the figure from below a projectile is launched from point labeled $A$ with an initial speed of $30\frac{m}{s}$ making an angle of $37^{\circ}$ (with the horizontal). An observer notices that the projectile hits an incline making an angle of $90^{\circ}$ as depicted in point labeled $B$. Find the flying time on seconds. (You may use $g=10\frac{m}{s^{2}}$)

The alternatives given are:

$\begin{array}{ll} 1.&1.2\,s\\ 2.&2.6\,s\\ 3.&3.4\,s\\ 4.&4.2\,s\\ 5.&5.2\,s\\ \end{array}$

So far the only thing which I could come up with was to establish that the equation for the projectile would be as follows:

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

Since the height isn't exactly given (i'm stuck) but for obvious reasons it cannot be zero. But I can fill the other terms.

$y(t)=y_{o}+v_{o}\sin37^{\circ} t-\frac{1}{2}gt^2$

(Note: For this part I'm assuming an approximation of the $37-53-90$ triangle being $3-4-5$.)

$y(t)=y_{o}+30\left(\frac{3}{5}\right) t-\frac{1}{2}\left(10\right)t^2$

$y(t)=y_{o}+18 t-5t^2$

But now what?.

I think for this part I'm missing a concept which I don't know how to use the given hitting angle. Can somebody help me with this?.

Answer 1

There are multiple equations relating the parameters of motion of a body under a constant acceleration. You've used the one whose standard form is $s=ut+\frac12 at^2$. But since you neither have nor need $s$ that's a dead end.

Try $v = u + at$ and don't forget the horizontal component of the motion.

Answer 2

You're not interested in the position at all, you're interested in the direction the projectile is going at a certain time. According to my physics textbook, if a projectile is fired with velocity $v_0$ with an angle of elevation $\theta$, then at time $t$:

$$v_x=v_0\cos\theta\\v_y=v_0\sin\theta-gt$$

Based on the diagram, we want the ending angle to be $-45^\circ=\tan^{-1}(-1)$, so we will set $-v_y=v_x$ in the above. (We will take advantage of the fact that $37^\circ$ is approximately the angle of a 3-4-5 right triangle to grossly simplify our trig.)

$$-v_0\cos\theta=v_0\sin\theta-gt\\-30(0.8)=30(0.6)-(10)t\\-42=-10t\\t=4.2\,\textrm s$$

Answer 3

hint

you start with an horizontal speed $v_h $ and a vertical speed $v_v$ that are easily determined.
$v_h$remains constant.
$v_v$ diminuishes by $gt$ from the initial value.
After a certain $t$ their ratio shall reach $tan(-45°)$ .

Thus ..

(since you asked for the solution)

$$ \eqalign{ & v_{\,h} = 30\cos \left( {37^ \circ } \right)\quad v_{\,v} = 30\sin \left( {37^ \circ } \right) - gt \cr & {{30\sin \left( {37^ \circ } \right) - gt} \over {30\cos \left( {37^ \circ } \right)}} = - \tan \left( {45^ \circ } \right) = - 1 \cr & t = {{30} \over g}\left( {\cos \left( {37^ \circ } \right) + \sin \left( {37^ \circ } \right)} \right) \approx 4.2 \cr} $$

Answer 4

in horizontal direction:

$x=v_0\cdot cos(\alpha)\cdot t\Rightarrow t=\frac{x}{v_0 \cdot cos(\alpha)}$

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

rearranging the terms of the quadratic in standard form:

$y=-\frac{g}{2\cdot v_0^2\cdot cos^2(\alpha)}\cdot x^2+tg(\alpha)\cdot x +y_0$

now the condition for the angle at point B is: $\frac{dy}{dx}=-1$

which by differentiating y leads to: $-\frac{g}{v_0^2cor^2(\alpha)}\cdot x+tg(\alpha)=-1$

which solves for x:

$x=\frac{1+tg(\alpha)}{g}\cdot v_0^2\cdot cos^2(\alpha)$

which further solves for t:

$t=\frac{1+tg(\alpha)}{g}\cdot v_0\cdot cos(\alpha)$

numerically: $t=\frac{sin(37^0)+cos(37^0)}{10}\cdot 30\approx 4.2$