A bomber flies horizontally with a speed of 431 m/s relative to the ground. The altitude of the bomber is 2630 m and the terrain is level. Neglect the effects of air resistance. The acceleration of gravity is 9.8 m/s^2.

Question

A) How far from the point vertically under the point of release does a bomb hit the ground? Answer in units of m

B.)At what angle from the vertical at the point of release must the telescopic bomb sight be set so that the bomb hits the target seen in the sight at the time of release? Answer in units of ◦

Answer 1

A. h = Vo*t + 0.5g*t^2 = 2630 m.
0 + 4.9t^2 = 2630
t^2 = 536.73
Tf = 23.2 s. = Fall time or time in air.

d = Xo * Tf = 431m/s * 23.2s. = 9985 m.

B. tanA = Y/X = 2630/9985 = 0.26340
A = 14.8o