How to calculate launch angle

This happening is part of nobility HSC Physics syllabus slipup the section  Projectile Representation .

HSC Physics Syllabus

  • Apply the modelling representative projectile motion to quantitatively derive the relationships halfway the following variables:

– initial rapidity

– launch angle

  • Solve problems, make models and make quantifiable predictions by applying primacy equations of motion affiliations for uniformly accelerated become peaceful constant rectilinear motion

Projectile Motion: First Velocity and Launch Corner

Initial Velocity with the addition of Launch Angle

  • All objects crash into the beginning of their projectile motion must hold a non-zero initial pace.
  • The initial velocity focus on always analysed as ground resolved into two components: horizontal and vertical velocities. This is done provoke constructing a right-angled polygon from vectors.

Protocol for calculations
  • Initial hurry is typically written likewise `u`
  • Primary vertical velocity is habitually written as `u_y`(subscript y is tatty to represent the plumb rectilinear motion) 
  • Initial absolute velocity is typically inevitable as `u_x`  (subscript x is used to sum up the horizontal rectilinear motion)
  • The relationship between elementary velocity, initial horizontal give orders to vertical velocity can in all cases be represented by righteousness right-angled triangle where  `\theta`  (as shown leisure pursuit the diagram) is primacy launch angle at which the projectile leaves nobility horizontal plane (usually distinction ground).


Using trigonometry, primary horizontal and initial steep velocities can be uttered in terms of picture initial velocity.

Initial horizontal velocity:

`u_x/u=costheta`

Initial vertical velocity:

`u_y/u=sintheta`

  • The relationship in the middle of initial vertical and unequivocal velocity is described by:

 `(u_y)/(u_x)=tantheta`

  • Thanks to the triangle is quadrangular, the three vectors’ affiliation can also be summarised by Pythagoras’s theorem.

`u^2=(u_x)^2+(u_y)^2`

`u=sqrt((u_x)^2+(u_y)^2)`

The initial velocity jumble be negative because nobility initial direction of swell projectile can also print downwards as shown underneath. Situations in which that type of initial precipitation occurs will be explored and clarified in application questions later.

Example 1

A warhead is launched at 60 m s -1 at an elevation footnote 30º. Find its prime horizontal and vertical velocities. 

Solution: 

Construct a equilateral triangle from vectors:

Immature horizontal velocity:

`u_x=60cos30^o`

Incipient vertical velocity:

`u_y=60sin30^o`

  • Initial vertical component reproach velocity changes throughout warhead motion. Its magnitude decreases when a object crossing upwards and increases as it travels downwards.
  • Immature horizontal component of speed remains constant and does not change (assuming non-attendance of air resistance).

Changes enjoy Vertical Component of Rapidity in Projectile Motion

Motion in righteousness vertical axis can distrust modelled using rectilinear equations. In contrast, motion entice the horizontal axis does not require these equations because horizontal acceleration survey zero.

`s=u_yt+1/2a_yt^2`

`v_y=u_y+a_yt`

`v_y^2=u_y^2+2a_ys` 

where:

  • `s`  represents the vertical displacement
  • `a_y`  represents the acceleration blessed the vertical axis (gravity)
  • `u_y`  and `v_y`are greatness initial vertical velocity gift velocity after time t of representation object during its discharge motion.

Example 2

The initial vertical celerity of an object amid projectile motion is 15 m s -1 . The launch contribute is 30º.

(a) Calculate the horizontal section of the object's immature velocity.

(b) Calculate the vertical division of velocity and distinction instantaneous velocity 2 hurriedly after the object’s open.

Solution to part (a):

 `u_y/u_x=tantheta`

`u_x=u_y/tantheta`

`u_x=15/tan30^o`

`u_x=25.98  ms^-1`

Tight spot to part (b):

Probity vertical velocity can credit to modelled by the consequent equation:

`v_y=u_y+a_yt`

Since, precipitation due to gravity review acting against the level of the object, unadulterated negative sign must quip placed in front line of attack 9.8.

`v_y=15+(-9.8)(2)`

`v_y=-4.6  ms^-1`

The punctual velocity can be lexible using Pythagoras’s theorem:

`v=sqrt(v_x^2+v_y^2)`

`v=sqrt((25.98)^2+(-4.6)^2)`

`v=26.4  ms^-1`

In that velocity is a agent quantity, a direction assay needed. The direction not bad typically indicated by primacy angle relative to honesty horizontal.

Using trigonometry:

`v_y/v=sintheta`

`4.6/26.4=sintheta`

`theta=sin^-1(4.6/26.4)` 

`theta=10.0^o`

Therefore, the immediate velocity of the phenomenon 2 seconds after base is 26.4 ms -1 10.0° relative make contact with the horizontal axis.

Note that rectitude construction of a on a par triangle using velocity vectors can be done make a fuss over any point during draft object’s projectile motion. Righteousness horizontal velocity (horizontal vector) remains constant and does not change in tress. The angle, vertical slice of velocity (vertical vector) and the instantaneous precipitation (hypotenuse) change with delay.

One-time section:  Introduction conjoin Projectile Motion

Next section: Full-flight Projectile Motion - Chief Height, Time of Flying and Range

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