For the projectile described (20 m/s at 30 degrees), what is the time of flight ignoring air resistance?

For the projectile described (20 m/s at 30 degrees), what is the time of flight ignoring air resistance?

• A. About 2.0 s
• B. About 1.0 s
• C. About 3.0 s
• D. About 4.0 s

Answer: (A)

In projectile motion with constant gravity and no air resistance, the total time a projectile stays in the air depends on its vertical motion. If it’s launched and lands at the same height, the time to rise to the top equals the time to come back down, so the flight time is twice the time to reach the peak. Compute the vertical component of the initial velocity: v0_y = 20 sin(30°) = 10 m/s. The time to reach the peak is t_up = v0_y / g ≈ 10 / 9.8 ≈ 1.02 s. Doubling this gives a total flight time of about 2.04 s, which is about 2.0 s. So the projectile is in the air for roughly 2.0 seconds. The other options would require a different vertical speed or a different launching/landing height.

In projectile motion with constant gravity and no air resistance, the total time a projectile stays in the air depends on its vertical motion. If it’s launched and lands at the same height, the time to rise to the top equals the time to come back down, so the flight time is twice the time to reach the peak.

Compute the vertical component of the initial velocity: v0_y = 20 sin(30°) = 10 m/s. The time to reach the peak is t_up = v0_y / g ≈ 10 / 9.8 ≈ 1.02 s. Doubling this gives a total flight time of about 2.04 s, which is about 2.0 s.

So the projectile is in the air for roughly 2.0 seconds. The other options would require a different vertical speed or a different launching/landing height.