Projectile Motion

Projectiles

Projectile Motion

Any object given an initial velocity and then left to move freely under gravity is a projectile. In projectile motion, the horizontal and vertical components of the object's motion are completely independent.
Projectiles follow a curved path because the horizontal velocity remains constant, while the vertical velocity is affected by the acceleration due to gravity, \(g\).

If something's projected at an angle (like, say, a javelin) you start off with both horizontal and vertical velocity. This can make solving problems trickier.
To solve this kind of problem, you need to use this method: Resolve the initial velocity into horizontal and vertical components.
Use the vertical component to work out how long it's in the air and/or how high it goes.
Use the horizontal component to work out how far it goes in the horizontal direction while it's in the air.

Projectile motion and air resistance

Air resistance (see page 1 45) causes a drag force that acts in the opposite direction to motion and affects the trajectory of a projectile.
The horizontal component of drag reduces the horizontal speed of the projectile, and reduces the horizontal distance the projectile can travel.
If the projectile has a vertical component of velocity, drag reduces the maximum height the projectile will reach, and steepens the angle of descent.

A gun fires a bullet at \(502\text{ }ms^{-1}\) horizontally. If the gun washeld \(1.61 m\) above the ground, how far will the bullet travel? Assume there's no air resistance and the ground is horizontal.

\(s=ut+\frac{1}{2}at^2\)
\(s=\frac{1}{2}gt^2\)
\(-1.61=\frac{1}{2}-9.81\times t^2\)
\(=0.5729\) seconds

\(502\times 0.5729=287.6m\)

A golf ball is hit at 12.1 ms\(^{-1}\) at an angle of 31.50\(^\circ\) above the horizontal from a podium 4.20 m above the surface of the ground. Calculate the maximum height above the ground reached by the ball. Assume there's no air resistance and the ground is horizontal.

\(u=\sin(31.5)\times12.1=6.322\)
\(v^2=u^2+2as\)
\(s=\frac{v^2-u^2}{2a}\)
\(s=\frac{-6.322^2}{2\times 9.81}=2.04\)
\(4.20+2.04=6.24m\)

Questions

Projectile Motion.
Resolve the motion first: horizontal and vertical.
Drag force is created that slows the projectile down.

Independent Motion

A cannonball is fired horizontally from a clifftop at 76.6m above the sea. The initial velocity is \(287ms^{-1}\) Calculate the time of flight and the horizontal distance travelled.

Vertical:
- \(s_{v}:-76.6\)
- \(u_{V}:\)
- \(v_{v}:\)
- \(a_{v}:-9.81\)
- \(t:\)
\(s=ut+\frac{1}{2}at^2\)
\(-76.6= 0+\frac{1}{2}\times 9.81 \times t^2\)
\(\sqrt{ \frac{2s}{a} }\)
\(t=\sqrt{ \frac{2\times -76.6}{-9.81} }=3.952\) seconds.

Horizontal:
- \(s_{h}:\)
- \(u_{h}:287\)
- \(v_{h}:287\)
- \(a_{h}:\)
- \(t:3.952\) (from previous step)
- \(s_{h}=v_{h}t\)
- \(287\times 3.952=1134.166\) metres.
- \(=1130\) metres.

Componants of Velocity

Velocity at ANY point is \(v=\sqrt{ {v_{x}}^2 +{v_{y}}^2}\)
At the highest point on the curve of the projectile, the only component is the vertical componant, this means the overall velocity will be slower at the peak of the curve.
Throughout the journey of the projectile, the velocity at the end is the same as the velocity at the start.
The angle of the projectile from the horizontal is \(\theta=\tan^{-1}\frac{v_{y}}{v_{x}}\)
On the downwards journey, you still measure the angle from the horizontal, but the angle is measured counter clockwise.

Measuring the Final Velocity

Vertical:
- \(s_{v}:-35.1\)
- \(u_{V}:0\)
- \(v_{v}:?\)
- \(a_{v}:-9.81\)
- \(t:?\)

Horizontal:
- \(s_{h}:?\)
- \(u_{h}:55.2\)
- \(v_{h}:55.2\)
- \(a_{h}:0\)
- \(t:?\)

\(v^2=u^2+2as\)
\(v^2=0^2+2\times -9.81 \times -35.1\)
\(v=26.24\)

Final Velocity:
- \(v=\sqrt{ {v_{x}}^2 +{v_{y}}^2}\)
- \(v=\sqrt{ 55.2^2+26.24^2 }=61.1ms^{-1}\)

Final Angle:
- \(\theta=\tan^{-1}\left( \frac{26.24}{55.2} \right)=25.4^\circ\)

Projectile Motion

Projectiles

Independent Motion

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