Physics-projectile motion-finding the angle

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An object is fired from a cliff 50m high, at an initial velocity of 100 m/s at an unknown angle. You have to find the angle required to fire the object to a bucket 10m away that is 2m tall.

I have tried to solve it but I just seem to be going in circles. I think there are too many variables. My teacher gave the skeleton of the question and I just imputed values to see if I could solve it. Is it actually possible to solve this question with the information given?

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4 Answers

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Let $\theta$ be the firing angle. The position of the projectile as a function of time is given by

\begin{align} x &= 100 \, t \cos(\theta) \\ y &= 100 \, t \sin(\theta) - \frac{g}{2} \, t^2 \enspace. \end{align} From the first equation we get $$ t = \frac{x}{100\cos(\theta)} \enspace. $$ Letting $x=10$ and substituting into the equation for $y$, one obtains

$$ 2-50 = 10 \tan(\theta) - \frac{g}{200\cos^2(\theta)} \enspace. $$

Recalling that $\frac{1}{\cos^2(\theta)} = 1+ \tan^2(\theta)$, one gets

$$ \tan^2(\theta) - \frac{2000}{g} \tan(\theta) - \frac{9600}{g} +1 = 0 \enspace. $$

Solving this quadratic equation for $\tan(\theta)$ with $g = 9.81$ and taking the arctangent of the two solutions yields $\theta_1 = 1.5660$ and $\theta_2 = -1.3606$. The first solution corresponds to shooting almost straight up and the second corresponds to aiming almost straight at the bucket. (Aiming straight at the bucket from the cliff top would lead to $\theta = -1.3654$.)

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Let, $\theta$ be the angle of projection with horizontal line then we have horizontal component $100\cos \theta$ & vertical component $100\sin \theta$ of velocity $100$ m/sec.

Let the $t$ be the time to hit the object then horizontal range $$=(100\cos \theta)(t)=100t\cos \theta$$ $$\implies 100t\cos \theta=10$$ $$t=\frac{1}{10\cos \theta}$$ In the same time $t$ projectile covers net $50-2=48\ m$ vetical height then we have $$h=u\sin \theta+\frac{1}{2}gt^2$$ Substituting the corresponding values we get $$48=-100\sin \theta+\frac{1}{2}(9.81)\left(\frac{1}{10\cos \theta}\right)^2$$ Can you proceed to solve for $\theta$?

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Since you already got somewhere I shall give tips so you can fill in the gaps. If anything not clear,shall get back.

$$ c = \cos \alpha ,s = \sin \alpha $$

Trajectory:

$$ \ddot y = -g ; \, \ddot x = 0 $$ Integrate

$$ \dot y = -g\cdot t + (V\cdot s); /, \dot x = (V\cdot c) $$

Again integrate

$$ y = -g\cdot t^2/2 + V\cdot s \cdot t ; /, x = V\cdot c \cdot t $$

With given x and y one can eliminate $t$ , solving a parabolic equation.

With $ x / t = ( V.c) $ one can find $ \alpha$.

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Construct the quadratic first.

We have $-4.9x^2 + 100x\sin(\theta)+50$. (since $100\sin(\theta)$ is upwards velocity)

So plugging in $10$ for $x$ and $f(x)=2$, we have $-440+1000\sin(\theta)=2$, or $1000\sin(\theta)=442$. This means that $\sin(\theta)=\frac{442}{1000}=-\frac{221}{500}$. Therefore $\theta = \sin^{-1}(\frac{221}{500})=0.457$ radians or $26.23$ degrees.

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