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Projectile motion formulae

Projectile motion formulae

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Summary

Projectile motion formulae

In a nutshell

It is possible to derive formulae for the time of flight, the horizontal range and the time taken to reach the greatest height. 



Formulae

Consider a particle projected at an angle on a horizontal plane:


Maths; Projectiles; KS5 Year 13; Projectile motion formulae


The following formulae can be derived:


​​Time of flight

Consider the vertical motion for the time the particle is in flight:

​uy=Usin⁡αs=0a=−gt=Tu_y=U\sin\alpha \quad s=0 \quad a=-g \quad t=Tuy​=Usinαs=0a=−gt=T​​


Substitute into s=ut+12at2s=ut+\dfrac{1}{2}at^2s=ut+21​at2 to give:​

​s=ut+12at20=(Usin⁡α)T−g2T20=T(Usin⁡α−gT2)T=0 or T=2Usin⁡αg\begin{aligned}s&=ut+\dfrac{1}{2}at^2\\\\0&=(U\sin\alpha)T-\dfrac{g}{2}T^2\\\\0&=T(U\sin\alpha-\dfrac{gT}{2})\\\\T&=0 \ or \ T =\dfrac{2U\sin\alpha}{g}\end{aligned}s00T​=ut+21​at2=(Usinα)T−2g​T2=T(Usinα−2gT​)=0 or T=g2Usinα​​​​

​​

Therefore:

Time of flight=2Usin⁡αg\boxed{\text{Time of flight}=\cfrac{2U\sin\alpha}{g}}Time of flight=g2Usinα​​​​


Horizontal range

Consider the horizontal motion:

​ux=Ucos⁡α s=Rt=T s=vtR=Ucos⁡α×T\begin{aligned} u_x&=U\cos\alpha \ \quad s=R \quad t=T\\\ \\s&=v t\\ R&=U\cos\alpha \times T \end{aligned}ux​ sR​=Ucosα s=Rt=T=vt=Ucosα×T​​​

​

Substitute in the time of flight T=2Usin⁡αgT=\cfrac{2U\sin\alpha}{g}T=g2Usinα​​​

​R=Ucos⁡α×2Usin⁡αg=U2×2sin⁡αcos⁡αg\begin{aligned}R&=U\cos\alpha\times\dfrac{2U\sin\alpha}{g} \\\\&=\dfrac{U^2 \times 2\sin\alpha\cos\alpha}{g}\end{aligned}R​=Ucosα×g2Usinα​=gU2×2sinαcosα​​​​


Substituting in using the double angle formula 2sin⁡αcos⁡α≡sin⁡2α2\sin\alpha\cos\alpha\equiv\sin2\alpha2sinαcosα≡sin2α gives the horizontal range as:​


​Range=U2sin⁡2αg\boxed{\text{Range}=\cfrac{U^2\sin2\alpha}{g}}Range=gU2sin2α​​​​


Greatest height

At the greatest height, the vertical component of the velocity is zero:

​u=Usin⁡αv=0a=−gt=Tu = U\sin \alpha \quad v=0 \quad a=-g \quad t=Tu=Usinαv=0a=−gt=T


Use v=u+atv=u+atv=u+at to give:

v=u+at0=Usin⁡α−gT\begin{aligned}v &=u + at \\0 &= U \sin\alpha - gT\end{aligned}v0​=u+at=Usinα−gT​​​


Rearrange for TTT to get the time to reach the greatest height:

​​

Time to reach greatest height=Usin⁡αg\boxed{\text{Time to reach greatest height}=\cfrac{U\sin\alpha}{g}}Time to reach greatest height=gUsinα​​​​​


Example

A particle is projected at 10 m.s−110\ m.s^{-1}10 m.s−1​ at 30°30\degree30° above the horizontal. Find the time of flight and horizontal range.

Time of flight=2Usin⁡αg=2×10×sin⁡309.8=1.02 s‾ (3 s.f.)\begin{aligned}\text{Time of flight}&=\dfrac{2U\sin\alpha}{g} \\\\&=\dfrac{2\times10\times\sin30}{9.8} \\\\&=\underline{1.02\ s} \ (3\thinspace s.f.)\end{aligned}Time of flight​=g2Usinα​=9.82×10×sin30​=1.02 s​ (3s.f.)​​​


Range=U2sin⁡2αg=102×sin⁡609.8=8.84 m‾ (3 s.f.)\begin{aligned}\text{Range}&=\cfrac{U^2\sin2\alpha}{g} \\\\&=\cfrac{10^2\times\sin60}{9.8} \\\\&=\underline{8.84\ m} \ (3 \thinspace s.f.)\end{aligned}Range​=gU2sin2α​=9.8102×sin60​=8.84 m​ (3s.f.)​​​



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FAQs - Frequently Asked Questions

How do I work out time of flight?

The formula for time of flight can be derived from using s=ut+1/2 at^2, rearranging for t.

How do you work out the formula for time taken to reach the greatest height for a projectile?

Use the fact that v=0, use v=u+at and rearrange for t.

What is the range of a projectile?

It is the maximum horizontal distance that a projectile travels before hitting the ground.

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