A projectile is fired over horizontal ground from the origin with an initial speed of \( 50 \mathrm{~m} / \mathrm{s} \). What firing angles will produce a range of 250 m ? Angles of \( \square \) degrees will produce of range of 250 m . Do not round until the final answer. Then round to two decimal places as needed. Use a comma to separate answers as needed.)

To find the firing angles that will produce a range of 250 m for a projectile launched with an initial speed of \( v_0 = 50 \, \text{m/s} \), we can use the range formula for projectile motion: \[ R = \frac{v_0^2 \sin(2\theta)}{g} \] where: - \( R \) is the range, - \( v_0 \) is the initial speed, - \( \theta \) is the launch angle, - \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)). Given that \( R = 250 \, \text{m} \) and \( v_0 = 50 \, \text{m/s} \), we can rearrange the formula to solve for \( \sin(2\theta) \): \[ 250 = \frac{(50)^2 \sin(2\theta)}{9.81} \] Calculating \( (50)^2 \): \[ (50)^2 = 2500 \] Now substituting this into the equation: \[ 250 = \frac{2500 \sin(2\theta)}{9.81} \] Multiplying both sides by \( 9.81 \): \[ 250 \cdot 9.81 = 2500 \sin(2\theta) \] Calculating \( 250 \cdot 9.81 \): \[ 2452.5 = 2500 \sin(2\theta) \] Now, divide both sides by 2500: \[ \sin(2\theta) = \frac{2452.5}{2500} \] Calculating the right side: \[ \sin(2\theta) = 0.981 \] Now we need to find \( 2\theta \): \[ 2\theta = \arcsin(0.981) \] Calculating \( \arcsin(0.981) \): \[ 2\theta \approx 78.46^\circ \] Now, divide by 2 to find \( \theta \): \[ \theta \approx \frac{78.46}{2} \approx 39.23^\circ \] Since the sine function is positive in the first and second quadrants, we also have: \[ 2\theta = 180^\circ - 78.46^\circ \approx 101.54^\circ \] Dividing by 2 gives: \[ \theta \approx \frac{101.54}{2} \approx 50.77^\circ \] Thus, the two angles that will produce a range of 250 m are approximately: \[ \theta \approx 39.23^\circ, 50.77^\circ \] Rounding to two decimal places, the final answer is: \[ \boxed{39.23, 50.77} \]