Projectile Motion: The x-value of an object moving under the principles of
projectile motion is x(theta, v_0, t) = (v_0*cos(theta))*t. A particular
projectile is fired with an intital velocity of v_0 = 250 ft/s and an angle of
elevation of theta = 60 degrees. It travels a distance of 375ft in 3 seconds.

Is the projectile more sensitive to errors in initial speed or angle of evelation?
• We are given the information below:

• $$v_0 = 250$$
• $$\theta = 60^{\circ}$$
• $$t = 3$$
• $$x(\theta, v_0, t) = (v_0cos(\theta))t$$
• To see which variable is sensitive to errors, we compute the total differential for 3 variables defined by

\begin{aligned} dz = x_\theta(\theta, v_0, t)d\theta + x_{v_0}(\theta, v_0, t)dv_0 + x_t(\theta, v_0, t)dt \end{aligned}

• Where the $$x_\theta, x_t$$ and, $$x_{v_0}$$ are the first order partial derivatives with respect to $$\theta$$, t and $$v_0$$ respectively. After computing the partial derivatives, we then plug in the numbers given to us and inspect which variables are subject to change.

\begin{aligned} dz = x_\theta(\theta, v_0, t)d\theta + x_{v_0}(\theta, v_0, t)dv_0 + x_t(\theta, v_0, t)dt \\ = -v_0sin(\theta)t \,d\theta + cos(\theta)t \,d{v_0} + v_0cos(\theta) \,dt \\ =-750sin(60^\circ) \,d\theta + 3cos(60^\circ) \,d{v_0} + 250cos(60^\circ) \,dt \\ =-375\sqrt{3} \,d\theta + 1.5 \,d{v_0} + 125 \,dt \end{aligned}

• To find whether the initial velocity $$v_0$$ or angle of elevation $$\theta$$ are the most sensitive, we notice that for small changes in $$d\theta$$, dz is greatly reduced whereas if $$dv_0$$ is changed in small increments, dz increases only slightly. Also note the coefficients of $$d\theta$$ and $$dv_0$$, $$d\theta$$ coefficient is about 433 times larger than $$dv_0$$ only in the case when the initial velocity is large.

• Thus the projectile is more sensitive to errors in angle of elevation than the initial velocity.