Aball 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?

If the speed is downwards, it is negative. Acceleration is gravity and is also negative. You should use the uniformly accelerated linear movement formula: y=(1/2)*a*t^2+vo*t+y0 Where y0=initial height, y=final height, t=time If final height=the ground then final height is 0 0=(1/2)*(-10m/s^2)*t^2+(-8.0m/s)*t+30.0m 0=(-5m/s^2)*t^2+(-8.0m/s)*t+30.0m Now you can use Bhaskara formula: a=-5 b=-8 c=30 t=(-b+-square root(b^2-4*a*c))/(2a) t=(-(-8)+-square root((-8)^2-4*(-5)*30))/(2*(-5)) t=-3.38 s-> IMPOSSIBLE t=1.78 s->OK