Question

A ball has to travel a distance of 162 m when launched at an angle of 32.9 degrees. Find the initial velocity. It returns to same level as when it was launched.

Answer 1

To find the initial velocity required for the ball to travel a horizontal distance (range) of 162 m when launched at an angle of 32.9 degrees, we can use the equation for projectile motion.

The range \( R \) of a projectile launched with an initial velocity \( v_0 \) at an angle \( \theta \) is given by the formula:

\[
R = \frac{v_0^2 \sin(2\theta)}{g}
\]

where:
- \( R \) is the range (162 m in this case),
- \( v_0 \) is the initial velocity,
- \( \theta \) is the launch angle (32.9 degrees),
- \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).

Rearranging the formula to solve for \( v_0 \), we get:

\[
v_0^2 = \frac{R \cdot g}{\sin(2\theta)}
\]

Taking the square root to find \( v_0 \):

\[
v_0 = \sqrt{\frac{R \cdot g}{\sin(2\theta)}}
\]

Now we can substitute the values:

1. Convert the launch angle \( \theta \) to radians for calculations:
- \( \theta = 32.9^\circ \).
- \( \sin(2\theta) = \sin(2 \times 32.9^\circ) = \sin(65.8^\circ) \).

Now, computing \( \sin(65.8^\circ) \):

Using a calculator:
- \( \sin(65.8^\circ) \approx 0.9063 \)

Now we can substitute into the equation:

- \( R = 162 \, \text{m} \)
- \( g = 9.81 \, \text{m/s}^2 \)

\[
v_0 = \sqrt{\frac{162 \cdot 9.81}{0.9063}}
\]

Calculating:

\[
v_0 = \sqrt{\frac{1580.62}{0.9063}} \approx \sqrt{1745.45} \approx 41.8 \, \text{m/s}
\]

Therefore, the initial velocity required for the ball to travel 162 m at a launch angle of 32.9 degrees is approximately **41.8 m/s**.