Projectile Motion

Projectile Motion Basics

Helena Dedic

• A particle is in projectile motion when the horizontal component of its velocity is different from zero and when the gravitational force is the only force acting on it.

◦ We will always choose a coordinate system with y-axis pointing upward and x-axis pointing along the horizontal. • Its acceleration is then equal to g and points downward. Therefore:

$a_x=0$ 

$a_y=-10m/s^2$

◦ The x-component of its velocity is constant. The $v_x-t$ graph is a horizontal straight line. Therefore:

$v_x(t)=v_{0x}$  

$\mathrm{\Delta }x(t)=v_{0x}t$  

◦ The y-component of its velocity is decreasing at a rate of 10 m/s every second. It is a linear function of time and its $v_y-t$ graph is a straight line with a negative slope of$-10m/s^2$. Therefore:

$v_y(t)=v_{0y}-10t$  

$\mathrm{\Delta }y(t)=v_{0y}t-5t^2$  

• The velocity at any time t is a vector:

$\overrightarrow{v}(t)=v_{0x}\hat{i}+(v_{0y}-10t)\hat{j}$ 

• The path of an object in projectile motion is a parabola. At the vertex of the parabola, the y-component of the velocity is zero and thus the velocity is:

$\overrightarrow{v}(t)=v_{0x}\hat{i}$ 

• When the vertical displacement is zero then the horizontal displacement is called range R. The range depends only on the initial speed of the projectile and on the direction of the velocity that is the angle $\theta$ above horizontal.

$R=\frac{v_i^{2}Sin2\theta }{g}$ 

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Derivation of the Equation for Horizontal Range

We begin with the notion that the velocity has a magnitude v0 and the direction  above horizontal. Then the component of the initial velocity are:

$v_{0x}=v_{0}Cos\theta$  

$v_{0y}=v_{0}Sin\theta$  

We also note that $\mathrm{\Delta }x(t)=R$ and $\mathrm{\Delta }y(t)=0$. We will substitute these values into the equations of kinematics for projectile motion:

$R=v_0(Cos\theta )t$ 

$v_y(t)=v_{0}Sin\theta -10t$ 

$0=v_0(Sin\theta )t-5t^2$ 

We isolate t from the third equation above:

$0=v_0(Sin\theta )$t- $5t$  

$t=\frac{v_{0}Sin\theta }{5}$  

and substitute for t in the first equation for the range:

$R=v_0(Cos\theta )\frac{v_{0}Sin\theta }{5}=\frac{v_0^{2}Sin\theta Cos\theta }{5}$ 

We note the trigonometric identity $sin2\theta =2sin\theta cos\theta$. We can substitute for $sin\theta cos\theta$ in the equation for the range and obtain:

$R=\frac{v_0^{2}Sin2\theta /2}{5}=\frac{v_0^{2}Sin2\theta }{10}$ 

So, in general:

$R=\frac{v_0^{2}Sin2\theta }{g}$

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Related Videoclips

1. Range of Projectile
See how the horizontal range of a projectile depends on the initial angle of incidence. The initial velocity is the same in each case.
2. Parabolic motion versus vertical free fall
Compare parabolic motion of a projectile to vertical free fall motion.
3. Projectile Motion: velocity vector
Velocity vector (along with its x and y components)as it changes in Projetile motion.
4. Monkey and a Gun
The monkey falls downward at the same rate as the bullet. The acceleration due to gravity is the same for both, even though the bullet is fired with a high initial velocity.

5. Lecture on Projectile Motion


6. Solving Projectile Motion Problems

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Simulations

Phet Projectile Motion Simulation

Circus Physics: Projectile motion

Exercises

• Exercises on Projectile Motion
  
  

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