I've tried to solve the following question from my workbook:

A bomb is being shot from a cannon with an initial velocity of 400m/s towards a target that's located 400m above it (vertical distance) and 15km from it (horizonal distance), at which angles (relatiely to the ground) the cannon can be positioned?

As part of my resolving attempts I composed these two equalities: \begin{align} 15000&=400\cos(\alpha)t\\ 400&=400\sin(\alpha)t-4.9t^2 \end{align} When $\alpha$ represents the angle of the cannon relatively to the ground and $t$ represents the time. These two equalities seem unsolveable to me :(

I'll be glad to know if it is solveable and if I thought about it right, if now what should I try differently?

We use these formulas for kinematics so if you'd like to offer a physical solution please use them: \begin{align}v&=v_0+at \\x&=x_0+v_0t+\frac{1}{2}at^2 \\x&=x_0+\frac{v_0+v}{2}t \\v^2&=v_0^2+2a(x-x_0) \end{align}


closed as off-topic by sammy gerbil, stafusa, Chris, Kyle Kanos, Jon Custer Jan 28 '18 at 16:11

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – sammy gerbil, stafusa, Chris, Kyle Kanos, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.


I'll begin with equation (2): $$ 400=400sin(\alpha)t-4.9t^2\\ 400 = 400t\sqrt{1-\cos^2(\alpha)}-4.9t^2\\ 400 = 400\sqrt{t^2-(\cos \alpha\cdot t)^2} - 4.9t^2 $$

But by equation (1) , $\cos \alpha \cdot t = \frac{15000}{400}=\frac{150}{4}=\frac{75}{2}=37.5$. Substituting that into eq. (2) gives

$$ 400 = 400\sqrt{t^2-7.5^2} - 4.9t^2 $$ Now you can define $u=t^2$ and we get $$ (400 + 4.9 u)^2 = 400^2 \cdot (u-7.5^2)$$

Does it look more solvable?

  • $\begingroup$ In other words, OP was on right track, but needed to apply a trigonometric identity. $\endgroup$ – Ben51 Jan 26 '18 at 17:48
  • $\begingroup$ I can not believe that I did not try this simple identity.. I tried much more intricate ones, such as double angle and sin/cos sum and difference. Thank you though, it teaches me that sometimes I have to think simple. $\endgroup$ – Ozk Jan 26 '18 at 19:30

Not the answer you're looking for? Browse other questions tagged or ask your own question.