For a projectile launched from ground level, neglecting air resistance, what is the time of flight T (assuming it lands at the same height)?

For a projectile launched from ground level, neglecting air resistance, what is the time of flight T (assuming it lands at the same height)?

• A. T = 2 v0 sin θ / g
• B. T = (v0^2 sin 2θ) / g
• C. T = 2 v0 cos θ / g
• D. T = v0 / (g sin θ)

Answer: (A)

The time the projectile stays in the air is governed by vertical motion under constant gravity. Its initial vertical speed is v0 sin θ, so it takes t_up = v0 sin θ / g to reach the highest point. Since gravity acts the same on the way down, it takes the same amount of time to return to the ground, giving a total flight time T = 2 v0 sin θ / g. This result comes from the vertical motion equation y(t) = v0 sin θ t − (1/2) g t^2, setting y = 0 for the landing (excluding t = 0) and solving for t. The horizontal motion doesn’t affect flight time when the launch and landing heights are equal. If the landing height were different, the full equation would be needed.

The time the projectile stays in the air is governed by vertical motion under constant gravity. Its initial vertical speed is v0 sin θ, so it takes t_up = v0 sin θ / g to reach the highest point. Since gravity acts the same on the way down, it takes the same amount of time to return to the ground, giving a total flight time T = 2 v0 sin θ / g. This result comes from the vertical motion equation y(t) = v0 sin θ t − (1/2) g t^2, setting y = 0 for the landing (excluding t = 0) and solving for t. The horizontal motion doesn’t affect flight time when the launch and landing heights are equal. If the landing height were different, the full equation would be needed.