Projectile Motion
A projectile is any object given an initial velocity, after which only gravity acts on it (neglecting air resistance). Despite its 2D path, it cleanly splits into independent horizontal and vertical motions.
Concept
The Key Idea
Horizontal and vertical motions are independent :
Horizontal: constant velocity (no horizontal force).
Vertical: constant downward acceleration g g g
Initial Conditions
For a projectile launched at speed u u u θ \theta θ
u x = u cos θ u_x = u\cos\theta u x = u cos θ u y = u sin θ u_y = u\sin\theta u y = u sin θ
Equations (taking origin at launch, up positive)
x ( t ) = u cos θ ⋅ t x(t) = u\cos\theta \cdot t x ( t ) = u cos θ ⋅ t y ( t ) = u sin θ ⋅ t − 1 2 g t 2 y(t) = u\sin\theta \cdot t - \tfrac{1}{2}g t^2 y ( t ) = u sin θ ⋅ t − 2 1 g t 2 v x = u cos θ , v y = u sin θ − g t . v_x = u\cos\theta,\qquad v_y = u\sin\theta - gt. v x = u cos θ , v y = u sin θ − g t .
Trajectory Equation
Eliminate t t t x = u cos θ t x = u\cos\theta\,t x = u cos θ t t = x / ( u cos θ ) t = x/(u\cos\theta) t = x / ( u cos θ ) y = x tan θ − g x 2 2 u 2 cos 2 θ . y = x\tan\theta - \frac{g\,x^2}{2u^2\cos^2\theta}. y = x tan θ − 2 u 2 c o s 2 θ g x 2 . parabola .
Key Results (flat ground)
Quantity Formula Time of flight T = 2 u sin θ g T = \dfrac{2u\sin\theta}{g} T = g 2 u sin θ Maximum height H = u 2 sin 2 θ 2 g H = \dfrac{u^2\sin^2\theta}{2g} H = 2 g u 2 sin 2 θ Range R = u 2 sin 2 θ g R = \dfrac{u^2\sin 2\theta}{g} R = g u 2 sin 2 θ Max range angle θ = 45 ° \theta = 45° θ = 45° Max range value R max = u 2 / g R_{\max} = u^2/g R m a x = u 2 / g
Derivations
Time of flight: When y = 0 y = 0 y = 0 u sin θ T − 1 2 g T 2 = 0 ⇒ T ( u sin θ − g T / 2 ) = 0 u\sin\theta\,T - \tfrac{1}{2}gT^2 = 0 \Rightarrow T(u\sin\theta - gT/2) = 0 u sin θ T − 2 1 g T 2 = 0 ⇒ T ( u sin θ − g T /2 ) = 0 T = 2 u sin θ / g T = 2u\sin\theta/g T = 2 u sin θ / g
Range: R = u cos θ ⋅ T = u cos θ ⋅ ( 2 u sin θ / g ) = u 2 sin 2 θ / g R = u\cos\theta \cdot T = u\cos\theta \cdot (2u\sin\theta/g) = u^2\sin 2\theta/g R = u cos θ ⋅ T = u cos θ ⋅ ( 2 u sin θ / g ) = u 2 sin 2 θ / g 2 sin θ cos θ = sin 2 θ 2\sin\theta\cos\theta = \sin 2\theta 2 sin θ cos θ = sin 2 θ
Max height: At apex, v y = 0 ⇒ u sin θ = g t peak v_y = 0 \Rightarrow u\sin\theta = gt_{\text{peak}} v y = 0 ⇒ u sin θ = g t peak t peak = u sin θ / g = T / 2 t_{\text{peak}} = u\sin\theta/g = T/2 t peak = u sin θ / g = T /2 H = u sin θ ( T / 2 ) − 1 2 g ( T / 2 ) 2 = u 2 sin 2 θ / ( 2 g ) H = u\sin\theta\,(T/2) - \tfrac{1}{2}g(T/2)^2 = u^2\sin^2\theta/(2g) H = u sin θ ( T /2 ) − 2 1 g ( T /2 ) 2 = u 2 sin 2 θ / ( 2 g )
Two Angles, Same Range
For a given speed, angles θ \theta θ 90 ° − θ 90° - \theta 90° − θ sin 2 θ = sin ( 180 ° − 2 θ ) \sin 2\theta = \sin(180° - 2\theta) sin 2 θ = sin ( 180° − 2 θ )
Worked Example
Q: A ball is kicked with speed 20 m/s 20\,\text{m/s} 20 m/s 30 ° 30° 30° g = 10 m/s 2 g = 10\,\text{m/s}^2 g = 10 m/s 2 T T T H H H R R R
Solution:
u x = 20 cos 30 ° = 10 3 ≈ 17.32 u_x = 20\cos 30° = 10\sqrt{3} \approx 17.32 u x = 20 cos 30° = 10 3 ≈ 17.32 u y = 20 sin 30 ° = 10 u_y = 20\sin 30° = 10 u y = 20 sin 30° = 10
T = 2 ( 10 ) / 10 = 2 T = 2(10)/10 = 2 T = 2 ( 10 ) /10 = 2 H = ( 10 ) 2 / ( 2 ⋅ 10 ) = 5 H = (10)^2/(2 \cdot 10) = 5 H = ( 10 ) 2 / ( 2 ⋅ 10 ) = 5 R = ( 20 ) 2 sin 60 ° / 10 = 400 ⋅ ( 3 / 2 ) / 10 = 20 3 ≈ 34.64 R = (20)^2 \sin 60°/10 = 400 \cdot (\sqrt{3}/2)/10 = 20\sqrt{3} \approx 34.64 R = ( 20 ) 2 sin 60°/10 = 400 ⋅ ( 3 /2 ) /10 = 20 3 ≈ 34.64
At landing, v x = 17.32 v_x = 17.32 v x = 17.32 v y = − 10 v_y = -10 v y = − 10 17.32 2 + 10 2 = 300 + 100 = 20 \sqrt{17.32^2 + 10^2} = \sqrt{300 + 100} = 20 17.3 2 2 + 1 0 2 = 300 + 100 = 20
Common Confusions
"Horizontal velocity changes during flight." — No, with no air resistance v x v_x v x
"Time of flight depends on horizontal velocity." — No, T T T u y u_y u y g g g
"At max height, the projectile is momentarily stationary." — Only the vertical velocity is zero; horizontal velocity continues.
Key Takeaways
Horizontal and vertical motions are independent.
Range = u 2 sin 2 θ / g = u^2\sin 2\theta/g = u 2 sin 2 θ / g θ = 45 ° \theta = 45° θ = 45°
Time of flight = 2 u sin θ / g = 2u\sin\theta/g = 2 u sin θ / g = u 2 sin 2 θ / ( 2 g ) = u^2\sin^2\theta/(2g) = u 2 sin 2 θ / ( 2 g )
Trajectory is a parabola.
Two complementary angles give the same range.