A bomber is flying horizontally over level terrain at a speed of 250 m/s relative to the ground and at an altitude of 3.20 km.
The bombardier releases one bomb. How far does the bomb travel horizontally between its release and its impact on the ground? Ignore the effects of air resistance.
m
(b) Firing from the people on the ground suddenly incapacitates the bombardier before he can call, "Bombs away!" Consequently, the pilot maintains the plane's original course, altitude, and speed through a storm of flak. Where is the plane relative to the bomb's point of impact when the bomb hits the ground?
behind the bomb
ahead of the bomb
directly above the bomb
(c) The plane has a telescopic bombsight set so that the bomb hits the target seen in the sight at the moment of release. At what angle from the vertical was the bombsight set?
To find the distance the bomb travels horizontally between its release and impact, we can use the equation of motion in the horizontal direction:
distance = speed × time
Since the speed of the bomber is given as 250 m/s and there is no acceleration in the horizontal direction, the distance traveled by the bomb is equal to the distance traveled by the bomber.
Given that the bomber is flying at a speed of 250 m/s, we need to find the time it takes for the bomb to hit the ground. To do this, we can use the equation of motion in the vertical direction:
final velocity^2 = initial velocity^2 + 2 × acceleration × distance
The initial velocity in the vertical direction is 0 m/s since the bomb is released from rest. The final velocity is the velocity of the bomb just before it hits the ground, which is also 0 m/s, since the bomb is at rest when it hits the ground. The acceleration due to gravity is approximately 9.8 m/s^2.
Using these values, we can solve for the distance traveled by the bomb vertically:
0^2 = 0^2 + 2 × 9.8 × distance
Using algebra, we can solve for the distance:
distance = 0.5 × 9.8 × distance
distance = 4.9 × distance
Simplifying further:
distance = distance
This equation tells us that the distance traveled by the bomb vertically is equal to the distance traveled by the bomb horizontally, which in turn is equal to the distance traveled by the bomber.
Therefore, the distance traveled by the bomb horizontally is 3.2 km.
(b) Since the pilot maintains the plane's original course, altitude, and speed through a storm of flak, we can assume that the position of the plane relative to the bomb's point of impact does not change. This means that when the bomb hits the ground, the plane will be directly above the bomb.
(c) The angle from the vertical at which the bombsight is set can be found using trigonometry. We can use the fact that the horizontal distance traveled by the bomb is equal to the distance traveled by the bomber.
The tangent of the angle from the vertical is given by:
tan(angle) = distance traveled horizontally / altitude
Substituting the given values:
tan(angle) = 3.2 km / 3.2 km
Since the numerator and denominator are equal, the tangent of the angle is 1, which means the angle is 45 degrees. Therefore, the bombsight is set at a 45-degree angle from the vertical.