A pilot wants to bomb his target on the ground. If the angle of the plane from the horizontal is 20° and the velocity of the plane is 80 m/s , find

A). The distance that the pilot must release the bomb to his target
B) the time before the bomb hits the target
C) the final velocity of the bomb before it hits the target.

V = 80m/s[-20O].

Vx = 80*Cos(-20) = 75.2 m/s.
Vy = 80*sin(-20) = -27.4 m/s.

A. Range = V^2*sin(2A)/g.
Range = 80^2*sin(2*20)/9.8 = 420 m.

B. Tan20 = h/420, h = 153 m.
h = Vy*t + 0.5g*t^2.
153 = -27.4*t + 4.9t^2,
4.9t^2 - 27.4t - 153 = 0,
Using the Quadratic Formula:
t = 9.04 s.

C. V = Vy + g*t = -27.4 + 9.8*9.04 = 61.2 m/s.

To find the answers to these questions, we can break down the problem into components: the horizontal and vertical components of the motion of the plane and the bomb.

A) The distance that the pilot must release the bomb to his target:
Since we want to find the horizontal distance, we need to consider the horizontal component of the motion. The horizontal component of the velocity is given by v_x = v × cos(θ), where v is the magnitude of the velocity (80 m/s) and θ is the angle of the plane from the horizontal (20°). Thus,

v_x = 80 m/s × cos(20°)
v_x ≈ 74.6 m/s

To find the distance, we can use the equation d = v × t, where d is the distance, v is the horizontal speed (v_x), and t is the time. We will find the time in the next step.

B) The time before the bomb hits the target:
To find the time, we need to consider the vertical component of the motion. The vertical component of the velocity is given by v_y = v × sin(θ), where v is the magnitude of the velocity (80 m/s) and θ is the angle of the plane from the horizontal (20°). Thus,

v_y = 80 m/s × sin(20°)
v_y ≈ 27.4 m/s

We can use the equation y = v_iy × t + (1/2)g × t^2, where y is the vertical displacement, v_iy is the initial vertical velocity (v_y), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Since the displacement in the vertical direction is zero at the target, we have:

0 = (v_y × t) - (1/2)g × t^2

Now, we can solve for t. Rearranging the equation gives us:

t × ((1/2)g) = v_y × t
(1/2)g × t = v_y
t = (2 × v_y) / g

Substituting the given values:

t = (2 × 27.4 m/s) / 9.8 m/s^2
t ≈ 5.6 s

C) The final velocity of the bomb before it hits the target:
Assuming air resistance is negligible, the final vertical velocity of the bomb just before hitting the target will be the negative of the initial vertical velocity:

v_fy = -v_y
v_fy ≈ -27.4 m/s

Therefore, the final velocity of the bomb before it hits the target is approximately -27.4 m/s in the vertical direction.

Note: It is important to verify the formulas, units, and assumptions used in the calculation to ensure accuracy when solving problems like this.