Question

A physics instructor has setup a projectile gun a horizontal distance d away from the target that is a height h above the ground. The instructor fires the projectile with an initial speed v0 at an angle θ0 with respect to the horizontal. At the instant the projectile leaves the gun, it is initially at the same height above the ground as the target. Ignore air resistance and take the gravitational acceleration g to be known.

a. Write down an vector expression for the initial velocity of the projectile. Clearly indicate your choice of coordinate system along with your choice of unit vectors.
b. Write down a vector expression for the position and velocity of the projectile when it has returned to the same height that is was fired out. Express your answers in terms of v0, θ0, and g.

Answer 1

a) $v(0)=(v_0 cos \theta_0) i + (v_0 sin \theta_0) j$ [m/s]

b)

$r=(\frac{v_0^2 sin(2\theta_0)}{g})i + (h)j$ [m]

$v=(v_0 cos \theta_0)i + (-v_0 sin \theta_0)j$ [m/s]

Explanation:

a)

The motion of the projectile consists of two independent motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration downward (acceleration due to gravity)

The data that we have are:

$v_0$ = initial speed of the projectile

$\theta_0$ = angle of projection with respect to the horizontal

$h$ = initial height of the projectile

$-g=$ gravitational acceleration

Here we take as coordinate system the positive x-axis in the forward direction and the positive y-axis in the upward direction, so the acceleration has a negative sign because it is downward.

The components of the initial velocity of the projectile along the x- and y- directions are:

$v_x = v_0 cos \theta_0\\v_y = v_0 sin \theta_0$

Therefore, in vector form,

$v(0)=(v_0 cos \theta_0) i + (v_0 sin \theta_0) j$ [m/s]

b)

The horizontal motion of the projectile is constant, so its horizontal velocity is constant, and the horizontal position at time t is

$x(t)=v_0 cos \theta_0 t$ (1)

Instead, the vertical motion is a uniformly accelerated motion, so vertical position is given by

$y(t)=h+v_0 sin \theta_0 t - \frac{1}{2}gt^2$ (2)

The particle reaches the target, which is located at same height of the initial height, when

$y(t)=h$

Substituting into (2), we can find the time t at which this happens, which is the time of flight:

$h=h+v_0 sin \theta_0 t - \frac{1}{2}gt^2\\v_0 sin \theta = \frac{1}{2}gt\\t=\frac{2v_0 sin \theta}{g}$ (3)

So, this is the time of flight. Substituting into (1), we also find the x-position when the projectile reaches the target:

$x=v_0 cos \theta_0 (\frac{2v_0 sin \theta}{g})=\frac{v_0^2 sin(2\theta_0)}{g}$

So, the position of the projectile at the target is:

$r=(\frac{v_0^2 sin(2\theta_0)}{g})i + (h)j$ [m]

Concerning the velocity, we know that the horizontal component of the velocity remains constant, so

$v_x = v_0 cos \theta_0$

The vertical velocity instead follows the law of uniformly accelerated motion, so:

$v_y(t) = v_0 sin \theta_0 - gt$

Substituting the expression for the time of flight, we find:

$v_y=v_0 sin \theta_0 - g(\frac{2 v_0 sin(\theta_0)}{g}) = v_0 sin \theta_0 - 2v_0 sin(\theta_0)=-v_0 sin \theta_0$

So, the velocity of the projectile at the target is

$v=(v_0 cos \theta_0)i + (-v_0 sin \theta_0)j$ [m/s]