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Physics · Ch 3 — Motion in a Plane

Projectile Motion

3.12

Projectile Motion

Projectile motion is the motion of an object launched into the air with some initial

velocity and then left to move freely under gravity alone, air resistance being neglected.

It is the most important application of two-dimensional motion with constant acceleration

in this chapter, and it is solved directly using the principle of independence of motion

(Section 3.11).

Setup. A projectile is launched from the ground with initial speed uu at an angle

θ\theta above the horizontal (see the figure). Its initial velocity resolves into a

horizontal component ux=ucos⁡θu_x = u\cos\theta and a vertical component uy=usin⁡θu_y = u\sin\theta. The

only acceleration acting is gravity, directed vertically downward: ax=0a_x = 0 (no horizontal

acceleration) and ay=−ga_y = -g (taking "up" as positive).

Horizontal motion therefore proceeds at constant velocity,

x(t)=(ucos⁡θ) t,x(t) = (u\cos\theta)\,t,

while vertical motion is uniformly accelerated, exactly like a ball thrown straight up:

y(t)=(usin⁡θ) t−12gt2,vy(t)=usin⁡θ−gt.y(t) = (u\sin\theta)\,t - \tfrac12 g t^2, \qquad v_y(t) = u\sin\theta - gt.

Trajectory. Eliminating tt between x(t)x(t) and y(t)y(t) gives the equation of the path:

y=xtan⁡θ−gx22u2cos⁡2θ,y = x\tan\theta - \frac{g x^2}{2u^2\cos^2\theta},

which is the equation of a parabola — confirming that every projectile (launched at any angle

other than exactly vertical) traces a parabolic arc.

Time of flight TT is the total time the projectile stays in the air, found by setting

y=0y = 0 (landing at the same height as launch):

T=2usin⁡θg.T = \frac{2u\sin\theta}{g}.

Maximum height HH is reached when vy=0v_y = 0 (the topmost point of the path), where the

velocity is purely horizontal, ucos⁡θu\cos\theta:

H=u2sin⁡2θ2g.H = \frac{u^2\sin^2\theta}{2g}.

Horizontal range RR is the horizontal distance covered in the full time of flight,

R=(ucos⁡θ) TR = (u\cos\theta)\,T, which simplifies to

R=u2sin⁡2θg.R = \frac{u^2\sin 2\theta}{g}.

Because sin⁡2θ\sin 2\theta reaches its maximum value of 1 when 2θ=90∘2\theta = 90^\circ, the range is

greatest at a launch angle of 45∘45^\circ, for a fixed launch speed uu. The table shows …

Figure 1Trajectory of a projectile launched at angle theta

What this figure shows. A projectile's parabolic trajectory launched from the origin with initial speed u at an angle theta above the horizontal ground. The velocity vector at launch is resolved into its horizontal component u cos(theta), shown as constant all along the path with dashed vertical guide lines marking equal horizontal steps in equal time intervals, and its vertical component u sin(theta), shown shrinking to zero at the topmost point of the path, marked as the point of maximum height H, then growing negative as the projectile falls back down. The horizontal ground distance from the launch point to the landing point is marked as the range R, and the total flight duration is marked as the time of flight T at the la …

Table 1Range and maximum height vs angle of projection (u = 20 m/s, g = 10 m/s²)
Angle of projection (θ\theta)sin⁡2θ\sin 2\thetaRange R=u2sin⁡2θ/gR = u^2\sin 2\theta / gMax height H=u2sin⁡2θ/2gH = u^2\sin^2\theta / 2g
15∘15^\circ0.50020.0 m1.34 m
30∘30^\circ0.86634.6 m5.0 m
45∘45^\circ1.00040.0 m10.0 m