Physics-projectile motion-finding the angle

Question

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?

Answer 1

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 $\theta$?

Answer 2

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$.

Answer 3

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.