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?

Respuesta :

W0lf93
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