Ballistic projectiles

A ballistic projectile is an object moving through the air under the force of gravity only. We give a projectile an initial speed by launching it; after that, it experiences acceleration due to gravity. The horizontal component of its velocity remains constant, but the vertical component decreases at a rate of 9.80 m/s2 (on Earth).

In our projectile problems, we assume that

All projectile problems can be solved by splitting up original velocity (v0\displaystyle \vec{v}_{0}) into its horizontal and vertical components with

vh=v0cosθandv1,v=v0sinθ\displaystyle \vec{v}_{\text{h}} = | \vec{v}_{0} \left \lvert \cos{\theta} \qquad \allowbreak\quad\allowbreak\text{and}\allowbreak\quad\allowbreak \qquad \vec{v}_{\text{1,v}} = \mid \vec{v}_{0} \right \rvert \sin{\theta},

where θ\displaystyle \theta is the direction of the original velocity expressed as an angle above horizontal. These can be used with the equations

Δdh=vhΔtandΔdv=v1,vΔt+12a(Δt)2\displaystyle \Delta{} \vec{d}_{\text{h}} = \vec{v}_{\text{h}} \Delta{} t \qquad \allowbreak\quad\allowbreak\text{and}\allowbreak\quad\allowbreak \qquad \Delta{} \vec{d}_{\text{v}} = \vec{v}_{\text{1,v}} \Delta{} t + \frac{1}{2} \vec{a} \left ( \Delta{} t \right )^{2}.

The duration of the projectile’s flight (Δt\displaystyle \Delta{} t) is associated with the vertical part. The range (distance to the landing point) is associated with the horizontal part.

Since we physics students don’t like doing any more work than absolutely necessary, we often use more specific, time-saving equations. The first equation tells us how long it takes (Δt\displaystyle \Delta{} t) for the projectile to return to its original altitude (Δdv=0\displaystyle \Delta{} \vec{d}_{\text{v}} = 0). If we let g=ag\displaystyle g{=} \left \lvert \vec{a}_{\text{g}} \right \rvert, then

Δt=2v0sinθg\displaystyle \Delta{} t = \frac{2 \left \lvert \vec{v}_{0} \right \rvert \sin{\theta}}{g},

and the range of this projectile is given by

Δdh=v02sin2θg\displaystyle \Delta{} \vec{d}_{\text{h}} = \frac{\left \lvert \vec{v}_{0} \right \rvert^{2} \sin{2} \theta}{g}.

The next equation doesn’t require the projectile to land at its original altitude. It treats the projectile’s altitude as a function of its horizontal position. This allows us to find the projectile’s vertical position at any particular horizontal position and, using the quadratic formula, go the other way as well. Assuming the projectile starts at the origin and passes through a point (x\displaystyle x, y\displaystyle y), then

y=(tanθ)xg2v02cos2θx2\displaystyle y = \left ( \tan{\theta} \right ) x - \frac{g}{2 \left \lvert \vec{v}_{0} \right \rvert^{2} \cos^{2}{\theta}} x^{2}.

The derivation of these three equations is left as an exercise to the reader.


A soccer ball is kicked at 14.5 m/s [29º a.h.]. Will it make it over a 1.5 m fence 5 m ahead?

When the ball has gone 5 m forward horizontally, x=5\displaystyle x = 5 and

y=(tan29º)(5m)9.80m/s22(14.5m/s)2cos2(29º)(5m)2=2.01m\displaystyle y = \left ( \tan{29} º \right ) \left ( 5 \, \text{m} \right ) - \frac{9.80 \, \text{m/s}^{2}}{2 \left ( 14.5 \, \text{m/s} \right )^{2} \cos^{2}{\left ( 29 º \right )}} \left ( 5 \, \text{m} \right )^{2} = 2.01 \, \text{m}.

Since 2.01m>1.5m\displaystyle 2.01 \, \text{m} > 1.5 \, \text{m}, the answer is yes—it will clear the fence.