A dive bomber has a velocity of 255 m/s at an angle θ below the horizontal. When the altitude of the aircraft is 2.15 km, it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.35 km. Find the angle θ.

Question

anonymous · Answer

If you do not consider the action of gravity, you will have a triangle:
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anonymous · Answer

Pretty sure i need to consider gravity as 9.8 m/s^2

anonymous · Answer

now, tan(theta) = 3.35/2.15
theta = atan(3.35/2.15)
theta = 1.00021rad = 57.31º

anonymous · Answer

Considering the gravity, we must first find the velocity components:
Vx = V*sin(theta)
Vy = V*cos(theta)

anonymous · Answer

vy = 30.188
vx = 26.242

anonymous · Answer

i got that far

anonymous · Answer

how you got these velocities?

anonymous · Answer

the same equations you jsut posted..

anonymous · Answer

that is not correct because 255 is not equal to sqrt(30.188^2 + 26.242^2)

anonymous · Answer

got it?

anonymous · Answer

L = Vx*t
L = V*sin(theta)*t, now find t in terms of theta

anonymous · Answer

H = Vy*t - 0.5*9.8*t^2
H = V*cos(theta)*t - 0.5*9.8*t^2

anonymous · Answer

solve for t, plug it back and solve for theta

anonymous · Answer

what is vx?

anonymous · Answer

is the x component of teh velocity