An owl is carrying a mouse to the chicks in its nest. It is 4.00 m west and 12.0 m above the center of the 30 cm diameter nest and is flying east at 4.50 m/s at an angle 35 below the horizontal when it accidentally drops the mouse. Will it fall into the nest? Find out by solving for the horizontal position of the mouse (measured in meters from the point of release) when it has fallen the 12.0 m

Can someone please check my work as my web assign keeps stating my answer as wrong, answer should be in sigfigs.

Expression for the radius of the nest is, r = D/ 2 --> Here, r is the radius of the nest and D is the diameter of the nest. Substitute 35 for D to find r. r = (35 cm 1m/100cm) / 2 = 0.18 r = 0.18 m horizontal component of the initial velocity of the owl is, Vox = Vo cos (A) Substitute 4.50 m/s for Vo and 35 for (A) to find Vox . Vox = 4.50 cos(35) Vox = 3.69 m/s vertical component of the initial velocity of the owl is, Voy = Vo sin (A) Substitute 4.50 m/s for Vo and 35 for (A) to find Vox . Voy = 4.50 sin(35) Voy = 2.58 m/s second kinematic equation is delta Y = Voy(t) + 1/2(g)(t)^2 Rearrange the above equation in the form of quadratic equation and substitute values 1/2(-9.80)(t)^2 + (-2.58)t - (-12.0) = 0 --> -4.9t^2 -2.58t + 12.0 = 0 t = -1.50 or 1.14s horizontal distance covered by the mouse is, delta X = Vox(t) delta X = (3.69 m/s) (1.14s) = The mouse falls 4.21 m away from the point of release.

Expression for the maximum distance to be covered by the mouse so that owl catches the mouse is, 4.00 - r = 4.00 - 0.18 = 3.82 or 4.18, No, the mouse misses the nest.