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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?

Ella
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4 Answers4

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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 $\theta = -1.3654$.)

Fabio Somenzi
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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 $\theta$?

Harish Chandra Rajpoot
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  • Why does the time it takes for the range to be covered be the same as the time it takes to drop the 48m to the bucket? – Ella Jul 30 '15 at 10:21
  • it's because the projectile is treated as a point (a single particle) which moves to cover the respective horizontal & vertical distances (simultaneously) in the same time. – Harish Chandra Rajpoot Jul 30 '15 at 10:25
  • But the object drops more than 48m because its path is parabolic, so doesn't that mean you can't use the value of 48 in your equation? – Ella Jul 30 '15 at 10:28
  • OK, remember the distance is the vertical height (same as the displacement) dropped it is not taken along the path. Kindly see I have edited it is a vertical height. – Harish Chandra Rajpoot Jul 30 '15 at 10:33
  • But aren't you saying that the time it takes for the projectile to reach 100xsin(theta) again, but in the opposite direction, is the same time it takes for the projectile to drop 48m into the bucket? – Ella Jul 30 '15 at 10:47
  • Kindly, notice the net vertical displacement of 48 m is in the downward direction that's why $g=9.81 \ m/sec^2$ is taken positive while $100\sin \theta$ is taken negative because it will automatically adjust the time of reaching to apex & that for returning to hit the object. In fact, you needn't to concern about time separately for going up & then returning down etc – Harish Chandra Rajpoot Jul 30 '15 at 10:56
  • The time it takes to cover the parabolic arch until it is at the height it was originally is added to the time it takes to drop the final 48m into the bucket. Thus you cannot say that it takes the same time. The time it takes to cover the 10m is including the whole flight not just the part where it falls the final 48m into the bucket. I cant get my head around what you're saying. – Ella Jul 30 '15 at 11:19
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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$.

Narasimham
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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.

Saketh Malyala
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