A ball is thrown off the edge of a building at an angle of \( 60^{\circ} \) and with the initial velocity of 5 meters per second. The equation that represents the horizontal distance of the ball x is \( \mathrm{x}=\mathrm{v}_0(\cos \theta) \mathrm{t} \) , where \( \mathrm{v}_0 \) is the initial velocity, \( \theta \) is the angle at which it is thrown and t is the time in seconds. About how far will the ball travel in 10 seconds?
Step-by-step Solution:
Step 1: Identify the Given Values - Initial velocity, \( v_0 = 5 \, \text{m/s} \), - Angle, \( \theta = 60^\circ \), - Time, \( t = 10 \, \text{s} \). Step 2: Calculate \( \cos \theta \) \[ \cos 60^\circ = \frac{1}{2} \] Step 3: Substitute Values into the Equation Substitute \( v_0 = 5 \), \( \cos \theta = \frac{1}{2} \), and \( t = 10 \) into the equation: \[ x = 5 \cdot \frac{1}{2} \cdot 10 \] Step 4: Simplify the Expression \[ x = 5 \cdot \frac{1}{2} \cdot 10 = 5 \cdot 5 = 25 \, \text{m} \] Step 5: Conclusion The ball will travel approximately: \[ \boxed{25 \, \text{m}} \] Correct Answer: \(\boxed{C}\)