FE Civil Dynamics Practice Problems

The current FE Civil specification assigns 4-6 of the 110 exam questions to Dynamics. Form A contains 4 original problems in this area. The sample below is published in full, with the same worked-solution format used throughout the book.

A water jet discharges from a nozzle at grade with an initial velocity of 18 m/s at an angle of $3 5^{\circ}$ above the horizontal. Air resistance is neglected. The horizontal distance from the nozzle to the point where the jet returns to nozzle elevation is most nearly:

(A)15.5 m
(B)18.9 m
(C)31.0 m
(D)37.9 m

Show worked solution

Answer: (C)

Symmetric flight over level ground: the jet stays aloft for $t = 2 v_{0} sin θ / g$ while drifting downrange at $v_{0} cos θ$ , landing 31.0 m from the nozzle.

Velocity components at launch:

v_{x} = 18 cos 3 5^{\circ} = 14.74 m/s

v_{y} = 18 sin 3 5^{\circ} = 10.32 m/s

Zero net vertical displacement fixes the time aloft:

t = \frac{2 v _{0} sin θ}{g} = \frac{2 ( 10.32 )}{9.81} = 2.105 s

Carried at constant $v_{x}$ for that time, the jet travels

x = v_{x} t = 14.74 \times 2.105 = 31.0 m

FE Reference Handbook — Dynamics: Projectile Motion

Why the other choices appear

(A)Omits the factor of 2 in the range expression, computing $v_{0}^{2} sin θ cos θ / g = 324 (0.5736) (0.8192) /9.81 = 15.5$ m.
(B)Uses $sin θ$ instead of $sin 2 θ$ in $R = v_{0}^{2} sin 2 θ / g$ , giving $324 (0.5736) /9.81 = 18.9$ m.
(D)Multiplies the full speed 18 m/s, instead of the horizontal component, by the flight time 2.105 s, giving 37.9 m.