Question

(1 point) Consider a cannonball launched along the path p(t)=(400t+80,200t−100,41t−16t2), with length units in feet and time units in seconds. When does it return to the height from which it was launched? (The coordinate system is righthanded, with z the vertical axis.) Answer: t= What is its velocity then? Answer: Momentum is mass times speed, usually measured in skgm, but here we can measure it in skgft. If the cannonball has mass 20 kg, what is its momentum upon impact with the ground? Answer: skgft

Answer 1

Error in the equation, but see below for the process.

Explanation:

p(t)=(400t+80,200t−100,41t−16t2

What is p? The problem says "path," but does that include the x, y, and z axes? Also, is the formula written correctly? I see two entrees of Number*t, and the last term, -16t2 also seems odd.

For this answer, we'll assume p(t) is the elevation (height) of the cannonball in feet at time x. I'll also assume the the final t is t^2, not t2.

p(t)=400t+80,200t−100,41t−16t^2

p(t)= t*(400 + 80,200 + 100,410)−16t^2

p(t) = 181010t - 16t^2

This seems an extremely odd equation, so please check.

We assume it is launched at 0 feet, ground level. We can solve the equation for p(t)=0. We know one solution will be at time 0. But solving the quadratic equation will give us the second time for which p(t) = 0.

p(t) = 181010t - 16t^2

0 = 181010t - 16t^2The two roots are t = 0 seconds (at launch) and t = (90505/8) or 11213 seconds

We can also graph the line and look for the intersection with the x axis. See the attached graph. The two points the line intersects the x axis are t = o and t = 11400 seconds.

Please check the equation, correct it, and repeat these steps.