The change in the linear momentum of the projectile is

\Delta \mathbf{p}=-\frac{50}{32.2}(1288) \mathbf{n}=-2000 \mathbf{n} \text { slug } \cdot \mathrm{ft} / \mathrm{sec},                (7.5a)

in which n is the unit normal vector directed into the wall. The average force  \mathbf{F}_S^*  acting on the projectile in the time  \Delta t=10^{-2}  sec is given by the last ratio in (7.4), and with (7.5a) we thus obtain

\mathbf{F}^* \equiv \frac{1}{\Delta t} \int_{t_0}^t \mathbf{F}(t) d t=\frac{\mathscr{G}\left(t ; t_0\right)}{\Delta t}=\frac{\Delta \mathbf{p}}{\Delta t}            (7.4)

\mathbf{F}_S^*=-\frac{2000}{10^{-2}} \mathbf{n}=-2 \times 10^5 \mathbf{n} \mathrm{lb}=-100 \mathbf{n} \text { tons }.                  (7.5b)

This estimates the total force exerted on the projectile by the concrete wall and gravity. Of course, the weight of the projectile compared with the total impulsive force (7.5b) is negligible, and hence the average force exerted on the wall by the projectile may be estimated by the equal and oppositely directed force  \mathbf{F}_W^*  = lOOn tons. If the action time increment is smaller, the average force acting on  the projectile or the penetration force acting on the wall grows larger.