To determine the angle of projection and the time of flight for a body given the conditions that its horizontal range is three times the maximum height and its initial velocity is 30 m/s, we can follow these steps:
### 1. Given Values:
- Horizontal Range [tex]\( R \)[/tex] is three times the maximum height [tex]\( H \)[/tex]: [tex]\( R = 3H \)[/tex]
- Initial velocity [tex]\( v = 30 \)[/tex] m/s
### 2. Fundamental Equations:
- Horizontal Range (R):
[tex]\[ R = \frac{v^2 \sin(2\theta)}{g} \][/tex]
- Maximum Height (H):
[tex]\[ H = \frac{v^2 \sin^2(\theta)}{2g} \][/tex]
### 3. Relationship Between Range and Height:
We are given [tex]\( R = 3H \)[/tex]. Substituting the expressions for [tex]\( R \)[/tex] and [tex]\( H \)[/tex]:
[tex]\[
\frac{v^2 \sin(2\theta)}{g} = 3 \left(\frac{v^2 \sin^2(\theta)}{2g}\right)
\][/tex]
### 4. Simplifying the Equation:
First, cancel the common terms ([tex]\( \frac{v^2}{g} \)[/tex]) on both sides:
[tex]\[
\sin(2\theta) = 3 \cdot \frac{\sin^2(\theta)}{2}
\][/tex]
Then we know from trigonometric identities:
[tex]\[
\sin(2\theta) = 2 \sin(\theta) \cos(\theta)
\][/tex]
So our equation becomes:
[tex]\[
2 \sin(\theta) \cos(\theta) = \frac{3}{2} \sin^2(\theta)
\][/tex]
Divide both sides by [tex]\( \sin(\theta) \)[/tex] (assuming [tex]\( \theta \)[/tex] is not zero):
[tex]\[
2 \cos(\theta) = \frac{3}{2} \sin(\theta)
\][/tex]
We rearrange to find [tex]\( \tan(\theta) \)[/tex]:
[tex]\[
\frac{\sin(\theta)}{\cos(\theta)} = \frac{4}{3}
\][/tex]
[tex]\[
\tan(\theta) = \frac{4}{3}
\][/tex]
### 5. Calculate the Angle of Projection:
To find [tex]\( \theta \)[/tex]:
[tex]\[
\theta = \tan^{-1} \left( \frac{4}{3} \right)
\][/tex]
This calculates to:
[tex]\[
\theta \approx 53.13^\circ
\][/tex]
### 6. Calculate the Time of Flight:
The formula for the time of flight [tex]\( T \)[/tex] is:
[tex]\[
T = \frac{2v \sin(\theta)}{g}
\][/tex]
Given [tex]\( v = 30 \)[/tex] m/s, [tex]\( \theta \approx 53.13^\circ \)[/tex], and [tex]\( g \approx 9.8 \)[/tex] m/s²:
[tex]\[
\sin(53.13^\circ) \approx 0.8
\][/tex]
[tex]\[
T = \frac{2 \cdot 30 \cdot 0.8}{9.8}
\][/tex]
[tex]\[
T \approx 4.90 \text{ seconds}
\][/tex]
### Final Results:
- The angle of projection is approximately [tex]\( 53.13^\circ \)[/tex].
- The time of flight is approximately [tex]\( 4.90 \)[/tex] seconds.
