GRE Physics Companion: Projectile Motion

This companion is designed for rapid review after M01-07. GRE style projectile questions are usually solved fastest by recognizing which assumptions allow a shortcut and which require the full component equations.

1 Fast triage

Always begin by resolving the launch velocity:

(1)

If launch and landing heights are equal, the shortcuts

(2)

and

(3)

are often fastest.

If the heights differ, solve

(4)

for the physical time, then use

(5)

2 Common traps

At the apex,

(6)

but neither the horizontal velocity nor the acceleration is generally zero.

The maximum same height range occurs at 45^∘ only in the ideal no drag model. Complementary angles give the same same height range, but different flight times and maximum heights.

For a horizontal launch, the time to fall depends only on the vertical drop, not on the horizontal launch speed.

If a projectile returns to the same height from which it was launched, its speed magnitude equals the launch speed in the ideal model, but its velocity vector does not.

3 Worked GRE example 1: compare complementary angles

Two projectiles are launched with the same speed from level ground at angles 30^∘ and 60^∘. Neglect drag. Compare their ranges and flight times.

The ranges are proportional to

(7)

For 30^∘,

(8)

For 60^∘,

(9)

Therefore the ranges are equal.

But

(10)

so

(11)

The 60^∘ projectile stays in the air longer.

4 Worked GRE example 2: horizontal launch ratio

Two balls roll horizontally from the same cliff at speeds v and 2v. They leave the edge simultaneously. Neglect air resistance. Compare their fall times and horizontal ranges.

The vertical initial velocity is zero for both, and both fall through the same height. Therefore their fall times are equal.

Since

(12)

and the second ball has twice the horizontal speed,

(13)

The faster horizontal motion changes range but not fall time.

5 GRE speed questions

1. A projectile is at the highest point of its trajectory. Which quantity must be zero? (A) v_x (B) v_y (C) a_x² + a _y² (D) speed
2. A projectile is launched and lands at the same height with fixed speed v₀. Which angle gives the maximum ideal range? (A) 30^∘ (B) 45^∘ (C) 60^∘ (D) 90^∘
3. Two ideal projectiles have equal launch speed and angles 25^∘ and 65^∘. Their same height ranges are (A) equal (B) in ratio 25∕65 (C) larger for 25^∘ (D) larger for 65^∘.
4. A ball is launched horizontally from a cliff. If its horizontal launch speed doubles, its fall time is (A) halved (B) unchanged (C) doubled (D) quadrupled.
5. A projectile returns to its launch height with no drag. Its speed magnitude just before return is (A) zero (B) less than v₀ (C) equal to v₀ (D) greater than v₀.
6. For an ideal projectile, the horizontal acceleration is (A) g (B) −g (C) zero (D) dependent on angle.
7. A projectile is launched from a platform and lands well below the launch height. Which method is safest? (A) always use R = v₀² sin 2𝜃∕g (B) solve the vertical equation for time, then use horizontal motion (C) set T = 2v_0y∕g (D) assume the path is symmetric about its apex and launch point.
8. At the same height on the upward and downward parts of an ideal trajectory, which statement is true? (A) the velocity vectors are identical (B) the speeds are equal (C) the vertical velocities have the same sign (D) the horizontal velocities have opposite signs.

6 Answers and rationales

1. B. At the apex v_y = 0; v_x and a_y = −g remain nonzero.
2. B. Same height range is proportional to sin 2𝜃, which is maximal at 2𝜃 = 90^∘.
3. A. The angles are complementary, so they give the same ideal same height range.
4. B. The fall time comes entirely from the vertical motion.
5. C. The projectile has recovered the same kinetic energy at the same height in the ideal model.
6. C. Ideal projectile motion has a_x = 0.
7. B. The standard same height shortcuts do not apply to unequal heights.
8. B. The speed magnitude is the same; the vertical component reverses sign.

References