A projectile is launched with speed 20 m/s at 45 degrees. What is the time of flight until it returns to the launch level (ignore air resistance)?

• A. approximately 2.88 s
• B. 1.5 s
• C. 3.2 s
• D. 4.0 s

Answer: (A)

The time of flight for a projectile landing at the same level depends on the vertical component of its initial velocity. With speed 20 m/s at 45°, the vertical component is v_y0 = 20 sin(45°) = 20(√2/2) = 10√2 ≈ 14.14 m/s. The motion is symmetric, so the total time in the air is twice the time to reach the peak: T = 2 v_y0 / g. Using g ≈ 9.8 m/s², T ≈ 2(14.14)/9.8 ≈ 28.28/9.8 ≈ 2.88 s. So the time of flight is about 2.88 seconds.