[6] 3. Let r(t) be the position of a projectile which is launched from the point (0,0,2) meters at t=0 sec with initial velocity (:50,60,80:)(m)/(sec). Assuming the projectile is influenced only by gravity, and that the acceleration due to gravity is 9.8(m)/(sec^(2)), we have r^('')(t)=(:0,0,-9.8:)(m)/(sec^(2)). Find the position function r(t), and

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[6] 3. Let r(t) be the position of a projectile which is launched from the point (0,0,2) meters at t=0 sec with initial velocity (:50,60,80:)(m)/(sec). Assuming the projectile is influenced only by gravity, and that the acceleration due to gravity is 9.8(m)/(sec^(2)), we have r^('')(t)=(:0,0,-9.8:)(m)/(sec^(2)). Find the position function r(t), and the time at which the projectile will hit the xy plane (i.e. the ground), to the nearest 0.01 second.

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Expert Answer to - [6] 3. Let r(t) be the position of a projectile which is launched from the point (0,0,2) meters at t

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Solution for - [6] 3. Let r(t) be the position of a projectile which is launched from the point (0,0,2) meters at t

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This an additional answer to - [6] 3. Let r(t) be the position of a projectile which is launched from the point (0,0,2) meters at t

(Solved): [6] 3. Let r(t) be the position of a projectile which is launched from the point (0,0,2) meters at t ...

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