A particle follows a path defined by y = 8-3t^2-5t. At what time is the particle at ground level?

Saturday, December 7, 2013

A particle follows a path defined by $\displaystyle{y}={8}-{3}{{t}}^{{2}}-{5}{t}$

The variable y represents the height of the particle for a given time t.

The particle is at ground level when y=0. For this problem, set y=0 and solve for t.

$\displaystyle{y}=-{3}{{t}}^{{2}}-{5}{t}+{8}$

$\displaystyle{0}=-{3}{{t}}^{{2}}-{5}{t}+{8}$

$\displaystyle{0}=-{\left({3}{{t}}^{{2}}+{5}{t}-{8}\right)}$

$\displaystyle{0}=-{\left({3}{t}+{8}\right)}{\left({t}-{1}\right)}$

$\displaystyle{t}={1},{t}=-\frac{{8}}{{3}}$

Since time may not be negative the final answer is t=1 second.

Final Answer: The particle is at ground level at 1 second.

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