Range, max height, time of flight — all derived: Question bank

See the parent for the full derivations: Range, max height, time of flight — all derived.

True or false — justify

Each line: a claim, then ::: then the verdict with reasoning.

Doubling the launch speed

u

doubles the range.

False. Range has

u^{2}

, so doubling

u

quadruples

R

(from

\frac{u ^{2} s i n 2 θ}{g}

). Speed matters far more than the linear intuition suggests.

Doubling

u

doubles the time of flight.

True.

T = \frac{2 u s i n θ}{g}

is linear in

u

, so twice the speed gives twice the airtime (for the same angle).

At the highest point the projectile's speed is zero.

False. Only the vertical velocity is zero at the top. The horizontal velocity

u cos θ

never changes, so the total speed there is

u cos θ

(not zero, unless launched straight up).

Firing at

3 0^{\circ}

and at

6 0^{\circ}

with the same speed gives the same range.

True. They are complementary (

3 0^{\circ} + 6 0^{\circ} = 9 0^{\circ}

), and

sin 2 θ = sin (18 0^{\circ} - 2 θ)

, so

sin 6 0^{\circ} = sin 12 0^{\circ}

. Same range, but the

6 0^{\circ}

shot flies higher and longer.

Those two complementary shots also spend the same time in the air.

False. Time depends on

sin θ

, not

sin 2 θ

. The steeper

6 0^{\circ}

shot has larger

sin θ

, so it stays up longer even though it lands at the same spot.

A projectile launched at

4 5^{\circ}

always reaches its maximum possible height.

False. Max range is at

4 5^{\circ}

; max height is at

9 0^{\circ}

(straight up), where

sin^{2} θ = 1

. At

4 5^{\circ}

the height is only

\frac{u ^{2}}{4 g}

, half of the straight-up maximum

\frac{u ^{2}}{2 g}

If you halve

g

(say, on the Moon-like world), the range doubles.

True.

R \propto \frac{1}{g}

, so halving

g

doubles

R

. Weaker gravity also doubles the airtime, which is why the ball travels farther.

The horizontal velocity decreases as the projectile rises.

False. No horizontal force acts (air ignored), so

u cos θ

stays constant for the entire flight. Only the vertical velocity changes.

The time to go up equals the time to come down.

True (flat ground, no air). Gravity is constant, so the vertical motion is a time-mirror: it decelerates going up and re-accelerates by the same amount coming down, symmetrically.

Maximum height is proportional to the square of the vertical launch speed.

True.

H = \frac{( u s i n θ ) ^{2}}{2 g} = \frac{u _{y}^{2}}{2 g}

. It cares only about the vertical component

u_{y} = u sin θ

; the horizontal part is irrelevant to how high it goes.

Spot the error

Each line states a flawed step; the reveal names the error and fixes it.

"Range

= u cos θ \times \frac{u s i n θ}{g}

The time used is only the rise time

\frac{u s i n θ}{g}

, not the full flight. Total time is

T = \frac{2 u s i n θ}{g}

— the factor of

2

is missing. Correct range doubles this to

\frac{u ^{2} s i n 2 θ}{g}

"For range, plug the angle straight in:

R = \frac{u ^{2} s i n θ}{g}

The product

2 sin θ cos θ

collapses to

sin 2 θ

, not

sin θ

. Using

sin θ

drops the

cos θ

factor entirely — it would wrongly predict maximum range at

9 0^{\circ}

"Max height:

H = \frac{u s i n ^{2} θ}{2 g}

The speed must be squared too:

H = \frac{u ^{2} s i n ^{2} θ}{2 g}

. It comes from

v_{y}^{2} = u_{y}^{2} - 2 g H

with

u_{y} = u sin θ

, so both the sine and the

u

get squared.

"A ball rolled off a cliff at

u = 20

m/s horizontal has range

\frac{u ^{2} s i n ( 2 \cdot 0 ^{\circ} )}{g} = 0

The flat-ground formula assumes launch and landing heights are equal. Off a cliff they differ, so you must re-derive

T

from the vertical drop

h = \frac{1}{2} g T^{2}

, then

R = u_{x} T

. The ball clearly does travel forward.

"At

θ = 9 0^{\circ}

the range is maximum because the ball flies for the longest time."

Long airtime is only half the story: at

9 0^{\circ}

the horizontal speed is

u cos 9 0^{\circ} = 0

, so

R = 0

. Range needs both airtime and forward speed — the balance point is

4 5^{\circ}

"Since

T

H

, and

R

all contain

sin

, they all vanish only at

θ = 9 0^{\circ}

They all vanish at

θ = 0^{\circ}

(thrown flat along the ground → no airtime).

R

also vanishes at

9 0^{\circ}

(no forward speed), while

T

and

H

are largest there. Different zeros for different quantities.

"Energy tells us max height depends on launch angle, so

H \propto sin θ

The energy argument gives

m g H = \frac{1}{2} m u_{y}^{2}

, so

H = \frac{u _{y}^{2}}{2 g} \propto sin^{2} θ

, not

sin θ

. It is the square of the vertical speed that converts to height.

Why questions

Why is time of flight independent of the horizontal launch speed?

Because the flight ends when the vertical motion returns to ground level, and vertical motion feels only gravity — the horizontal component neither speeds up nor slows down that clock.

Why does the range formula use

sin 2 θ

rather than

sin θ cos θ

directly?

They are equal: the identity

2 sin θ cos θ = sin 2 θ

just repackages the product of "goes up long enough" (

sin θ

) and "goes forward fast" (

cos θ

) into one clean expression whose peak at

4 5^{\circ}

is obvious.

Why does

4 5^{\circ}

maximise range but not height?

Range trades off airtime against forward speed, balanced when

2 θ = 9 0^{\circ}

i.e.

4 5^{\circ}

. Height ignores forward speed entirely and just wants maximal upward speed, which is largest at

9 0^{\circ}

Why can we treat horizontal and vertical motion separately at all?

The only force (gravity) points purely vertically, so it changes only the vertical velocity. The horizontal and vertical equations never share a force term — they are coupled solely through the shared time

t

Why does throwing straight up give zero range even though it stays airborne longest?

With

θ = 9 0^{\circ}

the horizontal component

u cos 9 0^{\circ} = 0

: there is nothing carrying it sideways, so it lands exactly where it left. Airtime without forward speed produces no horizontal distance.

Why is the peak reached at exactly half the total flight time?

Vertical velocity falls linearly from

u sin θ

0

(rising) then to

- u sin θ

(falling) under constant

g

. Zero velocity — the peak — sits at the midpoint of that symmetric linear graph, so

t_{peak} = T /2

Why does weaker gravity increase both the height and the range?

Smaller

g

means slower vertical deceleration, so the ball rises higher and stays up longer; the extra airtime, combined with unchanged horizontal speed, also carries it farther. Both

H

and

R

carry a

\frac{1}{g}

Edge cases

What happens to

T

H

R

θ = 0^{\circ}

All three are zero:

sin 0 = 0

. Launched perfectly horizontally from the ground, the ball is already at landing height, so it has no airtime and no rise — it never gets airborne.

What happens at

θ = 9 0^{\circ}

(straight up)?

T = \frac{2 u}{g}

(longest airtime) and

H = \frac{u ^{2}}{2 g}

(highest possible), but

R = 0

because

sin 18 0^{\circ} = 0

— no horizontal component to carry it anywhere.

What is the launch angle that gives maximum height for fixed speed?

θ = 9 0^{\circ}

H = \frac{u ^{2} s i n ^{2} θ}{2 g}

is largest when

sin^{2} θ = 1

. Straight up trades all the speed into vertical motion, giving

H_{max} = \frac{u ^{2}}{2 g}

Do the flat-ground formulas still hold if the ball lands on a raised platform?

No. They assume equal launch and landing heights. If the landing point is higher (or lower), you must re-solve the vertical equation

y = u sin θ t - \frac{1}{2} g t^{2}

for the actual final

y

, not

y = 0

What is the range if

θ

is just above

0^{\circ}

, like

1^{\circ}

Very small but nonzero:

R = \frac{u ^{2} s i n 2 ^{\circ}}{g}

. The ball barely leaves the ground and lands almost immediately — a nearly grazing shot with tiny airtime.

If two launches give the same range, must they share a launch speed?

Not necessarily — the traps here assume equal

u

, but a faster, steeper shot could match a slower, flatter one's range. Equal range from the same speed forces the angles to be complementary; different speeds break that link.

At the instant of landing (flat ground), how does the speed compare to the launch speed?

They are equal in magnitude. By time-symmetry the vertical speed regains

u sin θ

(now downward) while the horizontal speed stays

u cos θ

, so the landing speed is

u_{x}^{2} + u_{y}^{2} = u

Recall One-line self-test (click to reveal)

If someone tells you "range is maximised at $4 5^{\circ}$ , therefore height is too," what's the fastest rebuttal? ::: Height ignores forward speed and grows with $sin^{2} θ$ , peaking at $9 0^{\circ}$ ; the $4 5^{\circ}$ result is specifically the range trade-off between airtime and forward speed.

Connections

Range, max height, time of flight — all derived
Projectile Motion — components & independence
Equations of Motion (1D kinematics)
Vectors — resolving into components
Free Fall & Acceleration due to gravity
Trigonometric identities — double angle
Energy method — max height via conservation