Physics

Problem

31 A projectile is launched from level ground with an initial speed

v 0

at an angle

\theta

with the horizontal. If air resistance is negligible, how long will the projectile remain in the air? (A)

2 \mathrm{v}_{0} / \mathrm{g}

(B)

2 \mathrm{v}_{0} \cos \theta / \mathrm{g}

(C)

\mathrm{v}_{0} \cos \theta / \mathrm{g}

(D)

\mathrm{v}_{0} \sin \theta / \mathrm{g}

(E)

2 \mathrm{v}_{0} \sin \theta / \mathrm{g}

(A)

2 \mathrm{v}_{0} / \mathrm{g}

(B)

2 \mathrm{v}_{0} \cos \theta / \mathrm{g}

(C)

\mathrm{v}_{0} \cos \theta / \mathrm{g}

(D)

\mathrm{v}_{0} \sin \theta / \mathrm{g}

(E)

2 \mathrm{v}_{0} \sin \theta / \mathrm{g}

Intuitive explanation

The key is to focus on vertical motion only—horizontal velocity doesn't affect how long the projectile stays airborne. The vertical component $v_0 \sin \theta$ determines how high it goes and how long gravity takes to bring it back down to ground level. Use the kinematic equation for displacement with the vertical component of initial velocity and constant downward acceleration $g$ .

Intuitive explanation

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