MHB Calculating Elevation Angle to Hit Target at 100 m and 10 m

[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 = $$10m/s^2$$?

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

 

[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:

$$a_y=-g$$

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

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

For the horizontal component of motion, we have:

$$a_x=0$$

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

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

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

And so:

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

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

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

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

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

Arrange as:

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

Using a linear combination identity, we have:

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

Solving for $\theta$, there results:

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

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

 

[skeeter]

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 = $$10m/s^2$$?

$\Delta x = v_0\cos{\theta_0} \cdot t \implies t = \dfrac{\Delta x}{v_0\cos{\theta_0}} = \dfrac{100}{60\cos{\theta}}$

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

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 $\tan{\theta}$ ...

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

 

[MarkFL]

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

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

As a means of getting the other. :)

 

[Monoxdifly]

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:

$$a_y=-g$$

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

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

For the horizontal component of motion, we have:

$$a_x=0$$

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

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

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

And so:

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

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

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

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

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

Arrange as:

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

Using a linear combination identity, we have:

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

Solving for $\theta$, there results:

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

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?

 

[MarkFL]

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:

Linear Combination of Sine and Cosine