At a location where g = 9.80 m/s², an object is thrown vertically down with an initial speed of 1.00 m/s.After 5.00 s the object will have traveled: A)125 m B)128 m C)245 m D)250 m E)255 m

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Quizplus · Accepted Answer

The object's displacement can be calculated using the formula $s = ut + \frac{1}{2}gt^2$, where $u$ is the initial velocity, $g$ is the acceleration due to gravity, and $t$ is the time. Substituting $u = 1.00 \, \text{m/s}$, $g = 9.80 \, \text{m/s}^2$, and $t = 5.00 \, \text{s}$, we get $s = 1.00 \times 5.00 + \frac{1}{2} \times 9.80 \times (5.00)^2 = 5 + 122.5 = 127.5 \, \text{m}$, which rounds to 128 m.

At a Location Where G = 9