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Projection at any angle

Projection at any angle

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Explainer Video

Tutor: Meera

Summary

Projection at any angle

In a nutshell

Now that you have learnt how to resolve the initial velocity into horizontal and vertical components, it is possible to tackle projectile motion problems where the particle is projected at any angle. When considering the motion of a particle projected at an angle, the vertical and horizontal components are independent of each other, and therefore the calculations for vertical and horizontal movement can be considered separately. 



Projection at any angle

When a particle is projected at an initial velocity UUU at an angle α\alphaα above the horizontal, resolve the initial velocity into components. The horizontal and vertical motion can then be considered separately.


Horizontal motion

There is no acceleration in the horizontal direction as the only acceleration is vertical, (acceleration due to gravity). Therefore, the horizontal component of a projectile is considered to have constant velocity. The equation s=vt\bold s= \bold v ts=vt is used. The horizontal component of the initial velocity is Ucos⁡αU \cos \alphaUcosα.


​

Vertical motion

Acceleration due to gravity is 9.8 m.s−29.8 \ m.s^{-2}9.8 m.s−2, and so the formulae for constant acceleration can be used in the vertical direction. The vertical component of the initial velocity is Usin⁡αU \sin \alphaUsinα. When a projectile reaches its greatest height, the vertical component of the velocity is 000.​

​

Equations

description

equation

​Horizontal motion is modelled as having

constant velocity (a=0)(a=0)(a=0).​
​s=vts=vts=vt​​

​Vertical motion is modelled as having 

constant acceleration due to gravity, 

so the constant acceleration formulae can be used.
​v=u+ats=(u+v)2ts=ut+12at2s=vt−12at2v2=u2+2as\begin{aligned}v&=u+at \\ \\s&=\dfrac{(u+v)}{2}t \\\\s&=ut+\dfrac{1}{2}at^2 \\\\s&=vt-\dfrac{1}{2}at^2 \\\\v^2&=u^2+2as\end{aligned}vsssv2​=u+at=2(u+v)​t=ut+21​at2=vt−21​at2=u2+2as​​​


Variable definitions

​​quantity name

symbol

unit name

unit

​displacementdisplacementdisplacement​​
​sss​​
​metresmetresmetres​​
​mmm​​
​initial veloctyinitial \ veloctyinitial velocty​​
​uuu​​
​metres per secondmetres \ per \ secondmetres per second​​
​m.s−1m.s^{-1}m.s−1​​
​final velocityfinal \ velocityfinal velocity​​
​vvv​​
​metres per secondmetres \ per \ secondmetres per second​​
​m.s−1m.s^{-1}m.s−1​​
​accelerationaccelerationacceleration​​
​aaa​​
​metres per second squaredmetres \ per \ second \ squaredmetres per second squared​​
​m.s−2m.s^{-2}m.s−2​​
​timetimetime​​
​ttt​​
​secondssecondsseconds​​
​sss​​
​acceleration due to gravityacceleration \ due \ to \ gravityacceleration due to gravity​​
​g=9.8g = 9.8g=9.8​​
​metres per second squaredmetres \ per \ second \ squaredmetres per second squared​​
​m.s−2m.s^{-2}m.s−2​​


​

Example

A particle is projected from 3 m3\ m3 m​ above the horizontal ground with a velocity of (4i+6j) m.s−1(4i+6j)\ m.s^{-1}(4i+6j) m.s−1. Take the upwards and right each to be positive. Find:

i) The greatest height above the ground.

ii) The speed with which the particle hits the ground.

iii) The angle of the particle as it hits the ground.


i) To calculate the greatest height of the particle, consider the vertical motion of the particle:

s= ?u=6v=0a=−9.8t= ?s=\space? \qquad u = 6 \qquad v=0 \qquad a = -9.8 \qquad t=\space?s= ?u=6v=0a=−9.8t= ?​​

​v2=u2+2as02=62+2×(−9.8)×ss=1.84 m\begin{aligned}v^2&=u^2+2as\\0^2&=6^2+2\times (-9.8) \times s\\s&= 1.84\ m\end{aligned}v202s​=u2+2as=62+2×(−9.8)×s=1.84 m​​​

​

Since the particle was launched from 3 m3\ m3 m above the ground, the greatest height reached is:​

3+1.84=4.84 m‾ (3 s.f.)\underline{3+1.84=4.84\ m} \ (3\thinspace s.f.)3+1.84=4.84 m​ (3s.f.)


ii) The horizontal component of the velocity is always 4 m.s−14\ m.s^{-1}4 m.s−1. 


The vertical component can be calculated:

v2=u2+2asv2=62+2×(−9.8)×(−3)v2=94.8v=±9.7365 m.s−1\begin{aligned}v^2&=u^2+2as\\v^2&=6^2+2\times(-9.8)\times (-3)\\v^2&=94.8\\v&=\pm9.7365\ m.s^{-1} \end{aligned}v2v2v2v​=u2+2as=62+2×(−9.8)×(−3)=94.8=±9.7365 m.s−1​​​


The speed is therefore: 

v=42+9.73652=10.526 m.s−1v=\sqrt{4^2+9.7365^2}=10.526\ m.s^{-1}v=42+9.73652​=10.526 m.s−1


The particle hits the ground at 10.5 m.s−1 (3 s.f.)‾\underline{10.5\ m.s^{-1} \ (3 \thinspace s.f.)}10.5 m.s−1 (3s.f.)​.


iii) The angle of this collision is given by:

 tan⁡θ=9.73654 ⟹ θ=68°‾\tan{\theta}=\cfrac{9.7365}{4}\implies\theta=\underline{68\degree}tanθ=49.7365​⟹θ=68°​ (to the nearest degree)

​​


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FAQs - Frequently Asked Questions

How does the horizontal component of velocity change when a particle is projected at an angle?

The horizontal component of velocity does not change, it remains constant, assuming there is not air resistance.

When a particle is projected at an angle, what happens at the greatest height?

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

Can a particle be projected at any angle?

Yes, a particle can be projected horizontally, vertically or at any other angle.

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