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Calculus IIChapter 6: 3-Dimensional SpaceSection 6-11

Velocity and Acceleration — Question 2

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Question 2

A projectile is launched from ⟨0,0,5⟩\left\langle 0,0,5\right\rangle with initial velocity ⟨12,5,20⟩\left\langle 12,5,20\right\rangle ft/s. With gravity a→=⟨0,0,−32⟩\vec a=\left\langle 0,0,-32\right\rangle ft/s2^2, find its position, time of impact with z=0z=0, and horizontal range vector.

Original worksheet page 1: question and worked solution for 6-11-002
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Question 2 – Solution

Strategy Integrate constant acceleration and impose the ground condition on the vertical coordinate.

See the diagram in the original worksheet below.

Position r→(t)=⟨12t,5t,5+20t−16t2⟩.\vec r(t)=\left\langle 12t,5t,5+20t-16t^2\right\rangle.

Impact Solve 5+20t−16t2=05+20t-16t^2=0. The positive root is t*=5+458=5+358.\boxed{t_*=\frac{5+\sqrt{45}}8=\frac{5+3\sqrt 5}{8}}.

Horizontal range At impact, ⟨x(t*),y(t*)⟩=5+358⟨12,5⟩.\boxed{\left\langle x(t_*),y(t_*)\right\rangle =\frac{5+3\sqrt 5}{8}\left\langle 12,5\right\rangle}. The horizontal velocity is constant because gravity has no horizontal component.

Original worksheet page 2: question and worked solution for 6-11-002

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