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) 127.5 m C) 245 m D) 250 m E) 255 m

Question

Quizplus · Accepted Answer

The distance traveled by an object under constant acceleration can be calculated using the formula: $ s = ut + \frac{1}{2}at^2 $, where $ u $ is the initial velocity, $ a $ is the acceleration, and $ t $ is the time. Here, $ u = 1.00 \, \text{m/s} $, $ a = 9.80 \, \text{m/s}^2 $, and $ t = 5.00 \, \text{s} $. Plugging in these values, $ s = 1.00 \times 5.00 + \frac{1}{2} \times 9.80 \times (5.00)^2 = 5 + 122.5 = 127.5 \, \text{m} $.

At a Location Where G = 9