Projectile from a nozzle: time to reach the apex A water jet issues from a nozzle at 9.81 m/s making an elevation angle of 30° with the horizontal. Neglecting air resistance, how long does it take for a fluid particle to reach the highest point of its trajectory?

Introduction / Context:
Ideal projectile motion applies to fluid particles in a free jet exiting a nozzle, provided air resistance is neglected and the jet breaks into discrete particles. The time to reach the apex depends solely on the initial vertical component of velocity and gravitational acceleration.

Given Data / Assumptions:

• Initial speed u = 9.81 m/s at 30° above horizontal.
• Acceleration due to gravity g ≈ 9.81 m/s^2 downward.
• No air resistance; motion considered point-mass kinematics.
• Highest point is where vertical velocity equals zero.

Concept / Approach:
Time to the top is given by t_top = u_y / g, where u_y = u * sin θ is the vertical component of the initial velocity. At the apex, vertical velocity becomes zero due to gravity; horizontal motion does not affect the time to the highest point.

Step-by-Step Solution:

Verification / Alternative check:
Symmetry of projectile motion implies total flight time (to the same elevation) is 2 * t_top. If landing at nozzle height, total time would be 1.0 s, which is consistent with the computed half-time of 0.50 s.

Why Other Options Are Wrong:

• 0.25 s: Would require u_y = 2.45 m/s, not consistent with u = 9.81 m/s and 30° launch.
• 1.00 s and 1.50 s: These are total or longer-than-total times, not the ascent time.
• 2.00 s: Far too long given the vertical component and g.

Common Pitfalls:
Using the total time-of-flight formula instead of the ascent time; forgetting to multiply by sin θ when finding the vertical component; confusing degrees and radians for the angle.

Final Answer:
0.50 s