Projectile motion — horizontal - vertical independence, full derivation: Question bank

True or false — justify

A projectile's acceleration is zero at the top of its arc.

False — only the vertical velocity

v_{y}

is zero there; the acceleration is still

g

downward the entire flight, since gravity never switches off.

Horizontal velocity decreases as the projectile climbs.

False —

a_{x} = 0

, so

v_{x} = u cos θ

stays constant. The total speed drops because

v_{y}

shrinks, not because the sideways motion dies.

A heavier ball thrown at the same speed and angle has a shorter range.

False —

a_{y} = - g

contains no mass, so mass cancels; ignoring air drag, range depends only on

u

θ

g

, not on weight.

Doubling the launch speed doubles the range.

False —

R = \frac{u ^{2} sin 2 θ}{g}

scales with

u^{2}

, so doubling

u

quadruples the range.

Launching at

3 0^{\circ}

and

6 0^{\circ}

(same speed) gives the same range and the same time of flight.

Half true — ranges are equal (complementary angles), but

T \propto sin θ

, so the

6 0^{\circ}

shot stays airborne longer.

At the very top of the arc the projectile is momentarily at rest.

False — that is true only for a straight-up throw. In a general projectile

v_{x} = u cos θ \neq = 0

, so it is still cruising sideways at the top.

The time to reach maximum height equals the time to fall back to launch height.

True — vertical motion is symmetric about the top:

t_{up} = t_{down} = \frac{u sin θ}{g} = T /2

, because gravity is constant.

If you throw a ball horizontally off a table and drop another at the same instant, the thrown one lands later because it travels farther.

False — both share identical vertical motion (

u_{y} = 0

, same

g

, same height), so they land simultaneously; horizontal travel is irrelevant to landing time.

Increasing the launch angle always increases the range.

False — range grows only up to

4 5^{\circ}

, then falls again;

sin 2 θ

peaks at

2 θ = 9 0^{\circ}

The speed at landing (level ground) equals the launch speed.

True —

v_{x}

is unchanged and

∣ v_{y} ∣

at landing equals

u sin θ

by symmetry, so total speed

= u

(with air drag ignored).

Spot the error

"To find the range off a cliff, I'll use

R = \frac{u ^{2} sin 2 θ}{g}

Wrong — that formula assumes landing at launch height. For a cliff, solve the full

y = u_{y} t - \frac{1}{2} g t^{2}

for

t

, then

x = u_{x} t

"At the top,

v = 0

, so I set

u cos θ - g t = 0

to find that time."

Two errors:

v_{x}

u cos θ

and never zero, and gravity acts on

v_{y}

, not

v_{x}

. The top is where

v_{y} = u sin θ - g t = 0

"Since the ball slows as it rises, its horizontal distance per second shrinks too."

Wrong — only

v_{y}

shrinks;

v_{x}

is constant, so it covers equal horizontal distances in equal times throughout.

"The vertical equation is

y = u sin θ t + \frac{1}{2} g t^{2}

Sign error — up is positive and gravity pulls down, so it must be

- \frac{1}{2} g t^{2}

. The plus sign would make the ball rise forever.

"Time of flight is

T = \frac{2 u cos θ}{g}

Wrong component — flight time comes from vertical motion, so it uses

sin θ

T = \frac{2 u sin θ}{g}

"A projectile launched at

0^{\circ}

(horizontal) from the ground has zero range because

sin 0 = 0

The formula gives zero because it lands instantly at ground level — that is correct for level ground. Off a raised platform it does travel; the formula simply doesn't apply there.

Why questions

Why is the path a parabola and not a circle or straight line?

Because eliminating

t

gives

y = x tan θ - \frac{g x ^{2}}{2 u ^{2} cos ^{2} θ}

, which is quadratic in $x$ (

y = a x - b x^{2}

) — the defining shape of a parabola.

Why do horizontal and vertical motions not influence each other?

Newton's law holds per component and gravity only enters

a_{y}

; the two equations share nothing but the common clock

t

, so they are decoupled — see Newton's Second Law.

Why does

4 5^{\circ}

give the maximum range on level ground?

R \propto sin 2 θ

, and

sin 2 θ

is largest (

= 1

) when

2 θ = 9 0^{\circ}

, i.e.

θ = 4 5^{\circ}

— the sweet spot balancing horizontal reach against airtime.

Why does splitting the velocity use

cos θ

for horizontal and

sin θ

for vertical?

The velocity arrow's shadow on the x-axis (adjacent side) is

u cos θ

and on the y-axis (opposite side) is

u sin θ

— pure right-triangle geometry from Vectors — Resolving into Components.

Why do we integrate the acceleration to get velocity and position?

Because

a = \frac{d v}{d t}

and

v = \frac{d x}{d t}

— integrating undoes the derivative, turning known acceleration into velocity and then position (see Calculus — Integration).

Why is the landing speed's angle below horizontal equal to the launch angle above it (level ground)?

Vertical motion is symmetric, so

∣ v_{y} ∣

returns to

u sin θ

while

v_{x}

is unchanged; the velocity triangle is mirrored, giving the same angle magnitude.

Edge cases

What happens to the trajectory when

θ = 9 0^{\circ}

(straight up)?

u cos 9 0^{\circ} = 0

, so there is no horizontal motion — the "parabola" degenerates into a vertical line up and back down; range is zero.

What is the range when

θ = 0^{\circ}

launched from the ground?

Zero for level ground — the object is already at

y = 0

and never rises, so it "lands" at

t = 0

; a nonzero launch height is needed to see travel.

g

were doubled, what happens to time of flight, height, and range?

All shrink:

T \propto 1/ g

(halved),

H \propto 1/ g

(halved),

R \propto 1/ g

(halved) — stronger gravity yanks it down sooner. Compare Free Fall and g.

On the Moon (weaker

g

) at the same

u

and

θ

, how does the range change?

Range increases since

R \propto 1/ g

; weaker gravity means longer airtime and farther flight — same trajectory shape, just stretched.

What does the velocity vector look like exactly at the top of the arc?

Purely horizontal, of magnitude

u cos θ

; the vertical part has just crossed zero on its way from up to down, but the arrow never fully vanishes.

Is projectile motion still "independent" if there is a steady sideways wind?

No — wind adds a horizontal force, so

a_{x} \neq = 0

; the horizontal equation is no longer constant-velocity, though the vertical stays free fall. This ties into Relative Velocity.

Recall One-line summary to lock in

The only thing gravity touches is $v_{y}$ ; everything counterintuitive on this page comes from forgetting that $v_{x}$ is untouched and that time is set by the vertical axis alone.