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—¦
When you drop a bomb always turn and climb
The horizontal speed of the bomb is
THE SAME AS
the horizontal speed of the bomber
It does not change with no air drag !
The bomb hits UNDER the plane
so:
How long to fall?
4780 = (1/2)(9.8) t^2
t = 31.2 seconds to fall
how far horizontal?
175 * 31.2 = 5466 meters (Agree)
TAN THETA = 5466/4780
theta = 48.8 degrees from vertical
thnks for the help
:D
You are welcome.
To answer part a) of the question, you correctly used the formula for calculating distance (d) covered, which is given by d = v * t, where v is the velocity and t is the time of flight. However, there seems to be an error in your calculation. Let's go through the calculation step by step:
Given:
Velocity of the bomber (v) = 175 m/s
Altitude of the bomber (h) = 4780 m
Acceleration due to gravity (g) = 9.8 m/s^2
To find the time of flight (t), we can use the equation of motion for vertically upward motion:
h = ut - (1/2)gt^2
Substituting the known values:
4780 = 0 - (1/2)*(9.8)*t^2
Simplifying:
2.45t^2 = 4780
t^2 = 1951.02
t ≈ 44.16 seconds
Now, using the time of flight (t) and the horizontal velocity (v), we can calculate the horizontal distance (d) covered by the bomber:
d = v * t
d = 175 * 44.16
d ≈ 7716.88 m
Therefore, the bomb hits the ground approximately 7716.88 meters (or 7.72 kilometers) from the point vertically under the point of release.
Let's move on to part b) of the question. We need to find the angle (θ) from the vertical at the point of release where the bomb sight should be set to hit the target.
To solve this, we can use the fact that the horizontal and vertical motions are independent of each other. The horizontal velocity (v) remains constant throughout the motion, and the vertical motion is influenced only by gravity.
The tangent of the angle θ can be expressed as:
tan(θ) = vertical velocity / horizontal velocity
The vertical velocity (v_y) can be calculated using the formula:
v_y = g * t
Substituting the known values:
v_y = 9.8 * 44.16
v_y ≈ 432.768 m/s
Now, we can calculate the angle:
θ = arctan(v_y / v)
θ = arctan(432.768 / 175)
θ ≈ 68.39 degrees
Therefore, the telescopic bomb sight should be set at an angle of approximately 68.39 degrees from the vertical at the point of release to hit the target seen in the sight at the time of release.