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

• MHB
• Monoxdifly
In summary: Basically, we can express a sinusoidal function as a linear combination of sine and cosine functions with different amplitudes and phase shifts. In this case, we can use the identity to express the equation in terms of sine and cosine, making it easier to solve for $\theta$.
Monoxdifly
MHB
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.

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. :)

Monoxdifly said:
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 $\tan{\theta}$ ...

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

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. :)

MarkFL said:
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?

Monoxdifly said:
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

1. What is the formula for calculating elevation angle to hit a target at 100 m?

The formula for calculating elevation angle to hit a target at 100 m is: elevation angle = arctan (target height / horizontal distance).

2. How do I calculate the horizontal distance to my target?

The horizontal distance can be calculated using the Pythagorean theorem: horizontal distance = √(target distance² - target height²).

3. Can this formula be used for targets at different distances?

Yes, this formula can be used for targets at different distances. However, it may not be accurate for targets that are very far away or very close.

4. Do I need to know the exact height of my target for this calculation?

Yes, you will need to know the exact height of your target in order to calculate the elevation angle accurately.

5. How does the elevation angle affect the trajectory of my projectile?

The elevation angle determines the angle at which the projectile is launched, which affects its trajectory and the distance it will travel. A higher elevation angle will result in a longer distance, while a lower elevation angle will result in a shorter distance.

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