1. A target is located at the distance 100 m and height 10 m. How much is the elevation angle of the shooter to be able to hit the target if the initial velocity is 60 m/s and g = $\displaystyle 10m/s^2$?

What I have done:
100 = 60sin$\displaystyle \alpha$t and 10 = 60sin$\displaystyle \alpha$t - $\displaystyle \frac{1}{2}(10)t^2$
$\displaystyle t=\frac{5cos\alpha}{3}$ and $\displaystyle 10=t(60sin\alpha)-t)$
Dunno what to do from here on.

2.

3. Let's let $\theta$ be the angle of elevation. Let's orient our coordinate axes such that the vertical axis $y$ points up in the positive direction, and the horizontal axis $x$ points in the direction of motion in the positive direction.

For the vertical component of motion, we have:

$\displaystyle a_y=-g$

$\displaystyle v_y=-gt+v_0\sin(\theta)$

$\displaystyle y=-\frac{g}{2}t^2+v_0\sin(\theta)t$

For the horizontal component of motion, we have:

$\displaystyle a_x=0$

$\displaystyle v_x=v_0\cos(\theta)$

$\displaystyle x=v_0\cos(\theta)t$

Let's eliminate the parameter $t$, by using $\displaystyle t=\frac{x}{v_0\cos(\theta)}$:

And so:

$\displaystyle y=-\frac{g}{2}\left(\frac{x}{v_0\cos(\theta)}\right)^2+v_0\sin(\theta)\left(\frac{x}{v_0\cos(\theta)}\right)=-\frac{g}{2v_0^2\cos^2(\theta)}x^2+\tan(\theta)x$

Multiply through by $\cos^2(\theta)$:

$\displaystyle y\cos^2(\theta)=-\frac{g}{2v_0^2}x^2+\sin(\theta)\cos(\theta)x$

Using double-angle identities, and multiplying through by 2, we may write:

$\displaystyle y\left(\cos(2\theta)+1\right)=-\frac{g}{v_0^2}x^2+\sin(2\theta)x$

Arrange as:

$\displaystyle y+\frac{g}{v_0^2}x^2=\sin(2\theta)x-\cos(2\theta)y$

Using a linear combination identity, we have:

$\displaystyle \sqrt{x^2+y^2}\sin\left(2\theta-\arctan\left(\frac{y}{x}\right)\right)=y+\frac{g}{v_0^2}x^2$

Solving for $\theta$, there results:

$\displaystyle \theta=\frac{1}{2}\left(\arcsin\left(\frac{v_0^2y+gx^2}{v_0^2\sqrt{x^2+y^2}}\right)+\arctan\left(\frac{y}{x}\right)\right)$

Now, all that's left is to plug in the given values.

4. Originally Posted by Monoxdifly
A target is located at the distance 100 m and height 10 m. How much is the elevation angle of the shooter to be able to hit the target if the initial velocity is 60 m/s and g = $\displaystyle 10m/s^2$?
$\Delta x = v_0\cos{\theta_0} \cdot t \implies t = \dfrac{\Delta x}{v_0\cos{\theta_0}} = \dfrac{100}{60\cos{\theta}}$

$\Delta y = v_0\sin{\theta_0} \cdot t - \dfrac{1}{2}gt^2 \implies 10 = 60\sin{\theta} \cdot \dfrac{5}{3\cos{\theta}} - 5\left(\dfrac{5}{3\cos{\theta}}\right)^2 \implies 10 = 100\tan{\theta} - \dfrac{125}{9} \sec^2{\theta}$

using the identity $\sec^2{\theta} = 1 + \tan^2{\theta}$ yields a quadratic equation in $\tan{\theta}$ ...

$0 = -43 + 180\tan{\theta} - 25\tan^2{\theta}$

the quadratic formula yields two valid solutions for $\tan{\theta}$ ...

$\tan{\theta} = \dfrac{18 \pm \sqrt{281}}{5}$

5. Yes, there should be two solutions, and in my post above, I should have included:

$\displaystyle \sqrt{x^2+y^2}\sin\left(\pi-\left(2\theta-\arctan\left(\frac{y}{x}\right)\right)\right)=y+\frac{g}{v_0^2}x^2$

As a means of getting the other.

Originally Posted by MarkFL
Let's let $\theta$ be the angle of elevation. Let's orient our coordinate axes such that the vertical axis $y$ points up in the positive direction, and the horizontal axis $x$ points in the direction of motion in the positive direction.

For the vertical component of motion, we have:

$\displaystyle a_y=-g$

$\displaystyle v_y=-gt+v_0\sin(\theta)$

$\displaystyle y=-\frac{g}{2}t^2+v_0\sin(\theta)t$

For the horizontal component of motion, we have:

$\displaystyle a_x=0$

$\displaystyle v_x=v_0\cos(\theta)$

$\displaystyle x=v_0\cos(\theta)t$

Let's eliminate the parameter $t$, by using $\displaystyle t=\frac{x}{v_0\cos(\theta)}$:

And so:

$\displaystyle y=-\frac{g}{2}\left(\frac{x}{v_0\cos(\theta)}\right)^2+v_0\sin(\theta)\left(\frac{x}{v_0\cos(\theta)}\right)=-\frac{g}{2v_0^2\cos^2(\theta)}x^2+\tan(\theta)x$

Multiply through by $\cos^2(\theta)$:

$\displaystyle y\cos^2(\theta)=-\frac{g}{2v_0^2}x^2+\sin(\theta)\cos(\theta)x$

Using double-angle identities, and multiplying through by 2, we may write:

$\displaystyle y\left(\cos(2\theta)+1\right)=-\frac{g}{v_0^2}x^2+\sin(2\theta)x$

Arrange as:

$\displaystyle y+\frac{g}{v_0^2}x^2=\sin(2\theta)x-\cos(2\theta)y$

Using a linear combination identity, we have:

$\displaystyle \sqrt{x^2+y^2}\sin\left(2\theta-\arctan\left(\frac{y}{x}\right)\right)=y+\frac{g}{v_0^2}x^2$

Solving for $\theta$, there results:

$\displaystyle \theta=\frac{1}{2}\left(\arcsin\left(\frac{v_0^2y+gx^2}{v_0^2\sqrt{x^2+y^2}}\right)+\arctan\left(\frac{y}{x}\right)\right)$

Now, all that's left is to plug in the given values.
Thank you! All comprehensible except the linear combination identity part. What's that and how was it derived?

7. Originally Posted by Monoxdifly
Thank you! All comprehensible except the linear combination identity part. What's that and how was it derived?
Here's a link that describes the derivation of the identity: