PROJECTILE LAB

Hit the target

Pick angle and speed, swap gravity from Earth to the Moon, and split the path into its components.

Flight time

—s

measured on launch

Range

—m

measured on landing

Max height

—m

measured at the top

Speed now

15m/s

Earth · g = 9.81 m/s²

Mode

Launch angle45°

Launch speed15m/s

Platform height0m

Target distance25m

0 s

View

Target25 m

±0m

tolerance ±0.75 m

Setting up the scene…

Concepts

Two motions in one

A launched ball does two separate motions at once. Sideways there is no force, so it moves at constant velocity. Vertically only gravity acts, so it has constant acceleration

g

: it slows, stops rising and speeds up on the way down. The two motions never affect each other — the only thing they share is time.

x = v_x\,t \qquad y = h + v_y\,t - \tfrac12 g t^2

Splitting the launch velocity

The launch velocity is an arrow of length

v_0

at angle

\theta

. That arrow is the hypotenuse of a right triangle: the adjacent side is the horizontal component

v_x

, the opposite side the vertical component

v_y

. A steeper angle means more

v_y

and less

v_x

. In Components mode the blue arrow is

v_x

and the amber arrow is

v_y

v_x = v_0\cos\theta \qquad v_y = v_0\sin\theta \qquad v_0 = \sqrt{v_x^2 + v_y^2}

There is no "force of the throw"

The moment the ball leaves the barrel, the launcher stops pushing. The ball does not carry a "force of the throw"; in the air the only force is gravity (ignoring air). What carries the ball forward is not a force but its velocity — and since no force acts sideways,

v_x

never changes.

What happens at the top?

At the top only

v_y = 0

. The horizontal velocity

v_x

is still there, so the ball does not stop — it is just at its slowest (

v = v_0\cos\theta

). The acceleration is

g

downward at the top, exactly as everywhere else: gravity never switches off.

v_y = 0 \;\Rightarrow\; t = \frac{v_0\sin\theta}{g}, \quad v = v_0\cos\theta

Time of flight

Flight time is set only by the vertical motion. On level ground the climb and the fall are mirror images: the ball rises for

v_0\sin\theta/g

and falls for just as long. From a raised platform we solve

y(t) = 0

T = \dfrac{v_y + \sqrt{v_y^2 + 2gh}}{g}

T = \frac{2\,v_0\sin\theta}{g} \quad (h = 0)

Maximum height

The climb is exactly like a ball thrown straight up at

v_y

. At the top

v_y = 0

, so

v_y^2 = 2gH

H

grows with the square of

v_y

: double the vertical speed and the peak rises four times as high.

H = h + \frac{(v_0\sin\theta)^2}{2g}

Range

Range is horizontal speed times flight time:

R = v_x \cdot T

. On level ground this becomes

R = v_0^2\sin 2\theta/g

. Two big consequences: range grows with the square of the speed (2× speed → 4× range), and

\sin 2\theta

is largest at

\theta = 45^\circ

R = \frac{v_0^2\,\sin 2\theta}{g}

Does mass matter?

In a vacuum, no: gravity's pull is proportional to mass, so every object accelerates at

g

and heavy and light balls follow the same path. In air, yes: two balls of the same size feel the same drag force, but the light one's deceleration

a = F/m

is far larger. The foam ball slows quickly while the steel ball almost follows the ideal parabola.

a_{drag} = \frac{F_d}{m}

45° and complementary angles

\sin 2\theta

is the same for

\theta

and

90^\circ - \theta

, so 30° and 60° — or 15° and 75° — fired at the same speed land at the same spot. The steep shot stays up longer but crawls sideways; the two effects cancel exactly. 45° sits in the middle as the best balance.

\sin 2\theta = \sin(180° - 2\theta) = \sin 2(90° - \theta)

Dropped vs thrown

Two balls leaving the same height together — one dropped, one fired horizontally — land at the same time. Both start with

v_y = 0

and gain speed with the same

g

; sideways speed neither speeds up nor slows down the fall. Fall time depends only on height and

g

t = \sqrt{\frac{2h}{g}} \qquad x = u\,t

Reading a strobe picture

The faint "ghost" balls are snapshots taken at equal time steps

\Delta t

. Look at the floor shadow: equal gaps → constant horizontal velocity. Look at the wall shadow: the gaps shrink toward the top and grow on the way down → the vertical motion is accelerated.

Gravity on other worlds

Fire the same shot on another world and only

g

changes. Flight time, range and peak height **all scale with

1/g

**: on the Moon (

g = 1.62

m/s²) everything is about 6× Earth's, on Jupiter (

g = 24.79

m/s²) about 1/2.5. The shape of the path is identical — only its size changes.

T,\ R,\ H \;\propto\; \frac{1}{g}

Air resistance

In real air the ball feels a drag force opposite to its velocity. It grows with the square of the speed, so it bites hardest early on, while the ball is fast. Result: the range shrinks and the path becomes lopsided — a long, shallow climb and a short, steep fall. No closed formula exists; the lab integrates the motion numerically in small time steps (RK4).

F_d = \tfrac12\,\rho\,C_d\,A\,v^2

In air, 45° is not the best angle

With drag, high shots bleed away their speed over a long path. A flatter shot turns its horizontal speed into range more efficiently, so the best angle drops below 45°. In this lab, a baseball launched at 35 m/s does best at about 41°, and the light foam ball at the same speed at about 34°. The lab searches for this angle numerically every time you change a setting.

Skills and curriculum

Resolving the velocity vector

Grades 9–10

Split v₀ into vₓ = v₀·cos θ and vᵧ = v₀·sin θ.

Not started

MEB Grade 9 · Force and Motion · FİZ.9.2.4MEB Grade 10 · Geometric Shapes · MAT.10.1.1CCSS HSN-VM.A.3CCSS HSG-SRT.C.8

Independence of horizontal and vertical motion

Grades 10–11

Explain constant horizontal velocity and constant vertical acceleration, and why a dropped and a launched ball land together.

Not started

MEB Grade 11 · Force and Motion · FİZ.11.1.1, FİZ.11.1.3NGSS HS-PS2-1

Time of flight

Grades 10–11

Find time of flight from vertical motion: T = 2·v₀·sin θ / g (level ground).

Not started

MEB Grade 11 · Force and Motion · FİZ.11.1.3NGSS HS-PS2-1

Range and launch angle

Grades 10–11

Compute range; explain why 45° goes farthest and complementary angles share a range.

Not started

MEB Grade 11 · Force and Motion · FİZ.11.1.3NGSS HS-PS2-1

Maximum height

Grades 10–11

Use vᵧ = 0 at the top to find H = (v₀·sin θ)² / 2g.

Not started

MEB Grade 11 · Force and Motion · FİZ.11.1.3NGSS HS-PS2-1

Effect of gravitational acceleration

Grades 10–11

Explain how a smaller g on the Moon or Mars stretches time, range and height by 1/g.

Not started

MEB Grade 11 · Force and Motion · FİZ.11.1.1, FİZ.11.1.3NGSS HS-PS2-1

Air resistance

Grades 10–11

Explain how drag shortens range, skews the path and lowers the best angle below 45°.

Not started

MEB Grade 11 · Force and Motion · FİZ.11.1.8NGSS HS-PS2-1

Mission 1/16Explore0/16 ✓

First shots

Launch at at least three different angles and watch where each ball lands.

Try it in the simulation — the mission completes on its own when you reach the goal.

Hit the target

Concepts

Skills and curriculum

First shots

All missions

How it works: Hit the target

Keyboard and mouse