Questions: Two projectiles are launched from the ground simultaneously with an initial velocity 2.00 m / s. The first projectile is sent off at an angle of 16 with respect to the horizontal, and the second at an angle of 74. Which projectile lands on the ground in a region between 0.20 m and 0.23 m from the launcher?

Solution Steps

The horizontal 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 \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \).

For the first projectile, the initial velocity \( v_0 = 2.00 \, \text{m/s} \) and the launch angle \( \theta = 16^\circ \).

\[ R_1 = \frac{(2.00)^2 \sin(2 \times 16^\circ)}{9.81} \]

Calculate \( \sin(32^\circ) \):

\[ \sin(32^\circ) \approx 0.5299 \]

Substitute into the range formula:

\[ R_1 = \frac{4.00 \times 0.5299}{9.81} \approx 0.2163 \, \text{m} \]

For the second projectile, the initial velocity \( v_0 = 2.00 \, \text{m/s} \) and the launch angle \( \theta = 74^\circ \).

\[ R_2 = \frac{(2.00)^2 \sin(2 \times 74^\circ)}{9.81} \]

Calculate \( \sin(148^\circ) \):

\[ \sin(148^\circ) \approx 0.5299 \]

Substitute into the range formula:

\[ R_2 = \frac{4.00 \times 0.5299}{9.81} \approx 0.2163 \, \text{m} \]

Both projectiles have a calculated range of approximately \( 0.2163 \, \text{m} \), which falls within the specified region of \( 0.20 \, \text{m} \) to \( 0.23 \, \text{m} \).

Final Answer