Question

Question 1 The pilot is to bomb the abandoned structure in the picture. To stay safe from the explosion the plain is going to stay 1.83 km above the ground and launch the bomb at theta = 20 degrees . At what distance from the abandoned building should the mock bomb be released ? =5051km/h Vo Aban

Answer 1

g_km/s^2 = 9.8 times 10^-3 km/s^2 = 127008 km/ U_x = 505 cos 20 degrees = 474.54 km/h U_y = 505 sin 20 degrees = 172.72 km/h h = U^2 son^2 theta/2h = 0.117 km Total height = + 1.83 km = -1.947 km Total time = Time to read h + From h to ground = U sin theta/g + squareroot 2t_h/g = 172.72 km/h/127008 km/h^2 + squareroot 2((1.947)km/h/127008 = 0.00136 + 0.0055 = 0.00686 hr Thus, Distance, X_f = (474.54 km/hr)(0.00686 hr) = 3.255 km