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)`