A bomber flies horizontally with a speed of 175 m/s relative to the ground. The altitude of the bomber is 4780 m and the terrain is level. Neglect the effects of air resistance.
gravity is 9.8 m/s
a) How far from the point vertically under the point of release does a bomb hit the
ground?
I got 5465.25 for A.
by using 175 times 31.23 (the velocity times the time) I need help with B.
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 â—¦
a. 0.5g*t^2 = 4780.
4.9t^2 = 4780.
t^2 = 975.5
t = 31.2 s.
Dx = 175m/s * 31.2s = 5465.8 m.