In section 13.4 of Stewart's Calculus: Early Transcendentals, an example problem is worked out. I always seem to have problems grasping the physics stuff. In the example below I'll point out the spot where I can no longer follow.(adsbygoogle = window.adsbygoogle || []).push({});

(LaTeX codes don't seem to be working for me at the moment :grumpy: so I'll do this in ascii and revise with LaTeX when/if possible.)

A projectile is fired with an angle of elevation alpha and initial velocityv_0. Assuming that the air resistance is negligible and the only external force is due to gravity, find the possition function r(t) of the projectile. What value of alpha maximizes the range?

Solution: We set up the axes so that the projectile starts at the origin. Since the force due to gravity acts downward , we have

F= ma= -mgj

Where g = |a| ~= 9.8 m/s^2. Thus

a= -gj

Sincev'(t) =a, we have

v(t) = -gtj+C

WhereC=v(0) =v_0. Therefore

r'(t) =v(t) = -gtj+v_0

Integrating again, we obtain

r(t) = -1/2 gt^2j+ tv_0 +D

ButD=r(0) = 0, so the position vector of the projectile is given by

[3]r(t) = -1/2 gt^2j+ tv_0

***** I follow everything up to here fine, but the next step is where I get lost. *****

If we write |v_0| =v_0 (the initial speed of the projectile), then

v_0 =v_0 cos(alpha)i+v_0 sin(alpha)j

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# Homework Help: Example from 13.4, Stewart

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