Menu div.question,div.steps {font-size: 16px; padding: 10px;line-height: 20px; margin: 20px;} div.question{border: 1px solid #ccc;} div#solution .button{font-size: 20px; margin: 20px; width: 200px; padding: 10px; text-align: center;} div.hint,div.answer,div.steps{display: none;} div.pre_next{font-size: 30px; width:90%; margin:auto;} div.pre_next a.nav_pre{float:left;} div.pre_next a.nav_next{float:right;} div.title div.notes{ font-family:sans-serif, Arial; font-size: 18px; } 7. A ball is thrown directly downward with an initial speed of 8.00 m//s from a height of 30.0 m. After what time interval does it strike the ground? Hint y_f = y_i + v_it + 1/2 at^2 quadratic formula: ax^2 + bx + c = 0 than x = (-b ± sqrt(b^2 - 4ac))/(2a) Answer 1.79s Show Steps Givens: y_i = 30.0m y_f = 0 m v_i = -8.00 a=-g=-9.80m//s^2 Calculations: y_f = y_i + v_it +1/2 at^2 => 0 = 30.0 - 8.00t + 1/2(-9.80)t^2 => 4.90t^t +8.00t -30.0 =0 t = (-8.00 ± sqrt(8.00^2 - 4 times 4.90 times (-30.0)))/(2 times 4.90 Neglect the negative value we get: t = color(red)(1.79s)