To solve for the angle \(\theta\) of the tennis ball projected at an angle that achieves a range of \(R = 78 , m\) with an initial velocity \(v = 30 , m/s\), we can use the formula for the range of a projectile:
\[ R = \frac{v^2 \sin(2\theta)}{g} \]
where:
- \(R\) is the range,
- \(v\) is the initial velocity,
- \(g\) is the acceleration due to gravity.
Given:
- \(R = 78 , m\)
- \(v = 30 , m/s\)
- \(g = 10 , m/s^2\)
We can substitute these values into the formula and solve for \(\sin(2\theta)\):
\[ 78 = \frac{(30)^2 \sin(2\theta)}{10} \]
First, simplify the equation:
\[ 78 = \frac{900 \sin(2\theta)}{10} \]
\[ 78 = 90 \sin(2\theta) \]
Next, divide both sides by 90:
\[ \sin(2\theta) = \frac{78}{90} = \frac{13}{15} \]
Now, we need to find \(2\theta\):
To find the angle \(2\theta\), we can use the inverse sine function:
\[ 2\theta = \sin^{-1}\left(\frac{13}{15}\right) \]
Calculating \(2\theta\):
Using a calculator,
\[ 2\theta \approx 0.837 , \text{radians} \quad \text{(Convert to degrees if needed: } \approx 48.0^\circ\text{)} \]
Finally, to find \(\theta\):
\[ \theta = \frac{2\theta}{2} = \frac{0.837}{2} \approx 0.419 , \text{radians} \]
Convert to degrees:
\[ \theta \approx 0.419 \times \frac{180}{\pi} \approx 24.0^\circ \]
Thus, the angle \(\theta\) is approximately:
\[ \theta \approx 24^\circ \]