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?
`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)`
1.79s
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)`
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