Having just explored Newton's 3rd law in the context of the rocket principle, it is informative to continue to examine ballistic motion in this context. Thus, every rocket fired off at some defined angle Θ_o will behave with a motion such as that depicted. Note that in this motion, given the launch angle Θ_o, an apex will be reached where v(y) = 0 and also note the component v_x(o) remains the same throughout the flight. One can see from the geometry shown (and trig) that the relation of the initial velocities to launch angle are:

cos(Θ_o) = v_x(o)/ v(o) and sin (Θ_o) = v_y(o)/ v(o)

Then one can find the initial component velocities:

1) v_x(o) = v(o)cos(Θ_o) and (2) v_y(o)= v(o)sin (Θ_o)

Now, one also has kinematic equations to work with which define how bodies act undergoing motion. The two key equations are for velocity (in terms of initial velocity, u) and displacement (s) in terms of ut and ½ a t^2:

3) v = u + at = v(o) + at and (4) s = ut + ½ a t^2

Each of which now has to be modified for ballistic flight under the influence of gravity, so that 'g' must be reckoned in as the acceleration. But how to do it? First note that g must be matched to the velocity components. Since v_x(o) remains constant throughout the flight, there is no associated acceleration here. However, the component v_y(o) changes so that the acceleration must now be factored in. The component change in velocity, delta v_y(o)/delta t = a(y) = -g, so that our initial velocity equation is: v(y) = v_y(o) - gt, compared to v(x) = v(o)cos(Θ_o)

Now, v(y) = v_y(o) - gt but v_y(o)= v(o)sin (Θ_o) so that on substituting:

v(y) = v(o)sin (Θ_o) - gt

Now, we're in a position to obtain any x, y components during the ballistic flight. We have:

(5) x = u(x)t = v(o)cos(Θ_o)t and

(6) y = u(y)t + ½ a t^2 = v(o)sin (Θ_o)t - ½gt^2

And we're now in the position to solve some problems! Except for one more step that can make life even easier. That is, solve for t in equation (5) and then substitute this value into equation (6). On doing that we obtain:

y = tan(Θ_o)x - (g/ 2v(o)^2cos^2(Θ_o))x^2

which is valid for all angles from Θ_o = 0 to Θ_o = π/2. Note that this is the equation for a parabola (y = ax^2), which all ballistic paths follow when under the influence of gravity. There are three quantities in ballistic motion that can all be obtained from what we already have and are very critical:

a) the time t1, to reach maximum altitude

b) the value of the maximum altitude (call it h)

c) the

**range**or total distance covered, call it R.

The first can be obtained from our modified eqn. (3) or

v(y) = v(o)sin (Θ_o) - gt and solving for t, noting that v(y) = 0 at the apex of flight (see diagram):

0 = v(o)sin (Θ_o) - gt or t1 = v(o)sin (Θ_o)/ g

Now, substituting this value for t into equation (6) gives the maximum height:

y = h = v(o)^2sin^2 (Θ_o)/2 g

Which actual working is left as an exercise for the diligent reader or follower. Lastly, the range (R) is obtained by taking twice the time to cover reach the apex, or 2t1. It can be obtained by simply setting y = 0 in eqn. (6) then solving to get:

R = v(o)^2 sin (2Θ_o)/ g

This was assisted by making use of the trig identity: sin 2Θ = 2sinΘ cos Θ (also left as an exercise for readers). Now on to a problem:

A model rocket is launched at an angle of 60 degrees, at an initial velocity v(o) = 70 m/s. (Take g = 9.80 m/s/s) Find:

i) the time to reach maximum altitude

ii)the maximum altitude, h

iii) the total distance covered to the end of its flight.

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