Projectile Motion

• Calculate the horizontal and vertical components with respect to velocity and position of a projectile at various points along its path.

• Solve problems for projectiles launched horizontally and at various angles to the horizontal to calculate maximum height, range, and overall time of flight of the projectile.

Projectiles are the freely projected particles, which have the combined effect of a vertical as well as a horizontal motion.
Velocity of projection is defined as the velocity with which the particle is projected into space.
Angle of Projection ($\theta$) is the angle between the direction of projection and horizontal direction.
Trajectory is the path traced by the projectile.
\begin{aligned} v_x &= v_0 \cos \theta \\ v_y &= -gt + v_0 \sin \theta \end{aligned}

Time of flight is the time interval during which the projectile is in motion.
Time of flight, $t = \frac{2v_0 \sin\theta}{g}$

Maximum height of projectile, $h_{max} = \dfrac{v_0^2\sin ^2 \theta}{2g}$ Horizontal range is the horizontal distance through which the projectile travels during its flight.
Range = Maximum horizontal distance, $R= \dfrac{v_0^2}{g} \sin(2 \theta)$

Solved Example:

28-7-01

The horizontal range and maximum height of a projectile is same. The angle of projection is given by:

Solution:

$R = \dfrac{v_0^2}{g} \sin(2 \theta)$,

$h_{max} = \dfrac{v_0^2\sin ^2 \theta}{2g}$

But both are given equal, so equating,

\begin{align*} \dfrac{v_0^2}{g} \sin(2 \theta) &= \dfrac{v_0^2\sin ^2 \theta}{2g}\\ \sin 2\theta &= \dfrac{\sin^2\theta}{2}\\ 2 \sin \theta \cos \theta &= \dfrac{\sin^2\theta}{2}\\ \tan \theta &= 4 \end{align*}

Solved Example:

28-7-02

A man aimed his rifle at the bull's eye of a target 50 m away. If the speed of the bullet is 500 m/s, how far below the bull's eye does the bullet strikes the target?

Solution:
Horizontal motion: $t=\dfrac {50}{500}=0.1\ s$
Vertical motion: $s =ut +\dfrac{1}{2}at^{2} = \dfrac{1}{2}\left( 9.81\right) \left( 0.1\right)^{2} =0.049\ m\ \simeq 5\ cm$

Solved Example:

28-7-03

A baseball is thrown a horizontal plane following a parabolic path with an initial velocity of 100 m/s at an angle of 30$^\circ$ above the horizontal. Solve the distance from the throwing point that the ball attains its original level.

Solution:
Maximum range is given by: \begin{align*} R &= \dfrac{v^2}{g} \sin(2\theta)\\ &= \dfrac{100^2}{9.81} \sin(2 \times 30^\circ)\\ &=883\ m \end{align*}

Solved Example:

28-7-04

An archer must split the apple atop his partner's head from a distance of 30 m. The arrow is horizontal when aimed directly to the apple. At what angle must he aim in order to hit the apple with the arrow traveling at a speed of 35 m/s.

Solution:
Maximum range is given by: \begin{align*} R &= \dfrac{v^2}{g} \sin(2\theta)\\ 30 &= \dfrac{35^2}{9.81} \sin(2 \times \theta)\\ \end{align*} Solving, $\theta = 6.95^\circ$

Solved Example:

28-7-05

A projectile is fired from a cliff 300 m high with an initial velocity of 400 m/s. If the firing angle is 30$^\circ$ from the horizontal, compute the horizontal range of the projectile.

Solution:
Vertical motion: $s = ut+ \dfrac{1}{2}at^{2}$
$s = (-300) m,$
$u = 400 \sin 30^\circ\ m/s,$
$a = -9.81\ m/s^2$
Substituting, $4.905t^{2}-200t-300=0, \ \mathrm{Or}\ t=42.22\ sec$
Horizontal motion: \begin{align*} R &=\left( u\cos \theta \right) \cdot t\\ &=\left( 400\cos 30\right) 42\cdot 22\\ &=14626.5\ m\\ &=14.626\ km \end{align*}