# Secondary school Angular Throw/ Angular Motion [closed]

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 CusterJan 28 '18 at 16:11

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

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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 $$(400 + 4.9 u)^2 = 400^2 \cdot (u-7.5^2)$$