Question 7.14: Application to projectile motion. A projectile P is fired fr...

A simple free body diagram of the projectile will show that the only force acting on P is the conservative gravitational force mg = –mgj in frame φ. Consequently, the linear momentum in the horizontal direction i in φ, namely,  \mathbf{p} \cdot \mathbf{i}=m \dot{x},   is constant. Initially,   \mathbf{p} \cdot \mathbf{i}=m v_0 \cos \alpha;   hence , for all time,

\dot{x}=v_0 \cos \alpha.               (7.82a)

This easy result provides auxiliary information for later use.

Clearly, the system is conservative, and with y = 0 as the zero reference for the potential energy, the total energy initially is   E=\frac{1}{2} m v_0^2  At any subsequent position  , the potential energy is V(y) = mgy and the kinetic energy is  K(y)=\frac{1}{2} m v^2.  The energy principle (7.73) requires

K+V=E \text {, a constant. }              (7.73)

\frac{1}{2} m v^2  +  m g y=\frac{1}{2} m v_0^2,                   (7.82b)

which determines the projectile’s speed as a function of its altitude y :

v(y)=\sqrt{v_0^2  –  2 g y}.               (7.82c)

The projectile ‘s speed is independent of the gun’s angle of elevation and the mass of the projectile.

To find the greatest height attained, we recall that   v^2=\dot{x}^2  +  \dot{y}^2.  Clearly, the projectile attains its maximum altitude h when   \dot{y}=0.   With (7.82a), the speed   v(h)=\dot{x}=v_0 \cos \alpha,  and hence (7.82b) or (7.82c) yields the maximum altitude reached by P:

h=\frac{v_0^2}{2 g} \sin ^2 \alpha.           (7.82d)

Consequently, the greatest height attained depends on the angle of elevation, but not the mass of the projectile.

Finally, when P return s to the ground at Q, y = 0 and (7.82c) shows that the shell lands with speed equal to its muzzle speed  v_0.