I need help with this physics question please I cannot solve it. Please explain your work if you don't mind.

Question

A F-35 jet fighter moving horizontally at 200 m/s drops a missile, the missile immediately fires its engine, accelerating at 20 m/s^2 in the x-direction while it makes a free fall in the y-direction. When the missile has fallen 1000 m, find (a) its velocity in the y-direction, (b) its velocity in the x-direction, and (c) the magnitude and direction of its velocity. An old Mig-29 soviet fighter jet, 500 m ahead of the F-35 jet, was flying at 100 m/s, 1000 m below the F-35 fighter. (d) How much should the Mig-29 fighter jet accelerate to avoid being struck by the missile fired from the F-35? (e) if the maximum speed of the Mig-29 is Mach 1.5 can it escape the disaster?

Answer 1

Missile initial vertical velocity (vyi) = 0; initial horizontal velocity (vxi) = 200 m/s

Vertical drop distance (dy) = 1000 m; horizontal acceleration (ax) = 20 m/s²

Horizontal and vertical components are totally separate.

(a) How fast vertically (vyf) after falling 1000 m?

vyf² = vyi² + 2gd = 0 + 2(9.8)(1000) = 19600; vyf = 140 m/s

(b) How fast horizontally (vxf) after falling 1000 m?

Find time (t) of fall first; dy = 0.5gt²; 1000 = 4.9t²; t = 14.3 s

vxf = vxi + axt = 200 + (20)(14.6) = 485.7 m/s

(c) Velocity vector (v) after falling and accelerating? v² = vxf² + vyf² = 485.7² + 140² ; v = 505.5 m/s

tan θ = vyf / vxf = 140/485.7 = 0.288; tan^-1 0.288 = 16.1°

v is 505.5 m/s at 16.1° below horizontal

(d)How much acceleration (amig) will keep the Mig ahead of the missile?

First, find the distance the missile travels horizontally (dx) as it falls 1000 m. vxf² = vxi² + 2axdx;

485.7² = 0 + 2(20)dx; dx = 5897.6 m

Mig starts 500 m horizontally (dmigi) ahead of the missile, moving at 100 m/s (vmigi) and must be more than 5897.6 m ahead after 14.3 seconds to be safe: dx = dmigi + vmigit + 0.5at²

5897.6 = 500 + (100)(14.3) + 0.5amig(14.3)² ; amig = 17 m/s² is the minimum acceleration needed

(e) Is the Mig's speed (vmigf) after 14.3 seconds (at the minimum acceleration) less than Mach 1.5, that is 1.5 times the speed of sound (340 m/s). Mach 1.5 is 510 m/s.

vmigf² = vmigi² + 2amigd = 100² + 2(17)(5897.6) = 210518.4; vmigf = 458.8 m/s therefore it can escape

Answer 2

Let's first define the values given both given to us and what is implied:

v_0,x = 200 m/s

v_0,y = 0 m/s

a_x = 20 m/s²

a_y = -9.81 m/s²

y_f = -1000 m

y₀ = 0 m

(a) Find v_f,y using the following kinematic equation:

v_f,y = v_o,y + a_yt

All of these terms are defined except for time "t". We need to find out how long it takes for the missile to fall 1,000 m. Let's use this kinematic equation:

y_f,y = y_0,y + v_o,yt + 0.5a_yt²

-1000 = 0.5*(-9.81)*t²

t = 14.28 seconds

Use this time to find the final velocity in the y-direction:

v_f,y = (-9.81)*14.28

v_f,y = -140.09 m/s

(b) Velocity in the x-direction

v_f,x = v_o,x + a_xt

v_f,x = 200 + 20*14.28

v_f,x = 485.60 m/s

(c) Magnitude and direction of its velocity.

To find magnitude, use the following equation: v_f = √(v_f,x² + v_f,y²)

v_f = 505.40 m/s

To find direction, use the following equation: θ = tan^-1(v_f,y / v_f,x)

θ = -16.09° = 343.91°

(d) Let's define the values for the Mig-29 soviet plant

x₀ = 500 m ahead of the missile

v_0,x = 100 m/s

y₀ = 1000 m below the plane

It takes 14.28 seconds for the missile to reach the same elevation as the Mig-29. Let's determine the amount of distance in the x-direction the missile went.

x_f,x = x_0,x + v_0,xt + 0.5a_xt²

x_f,x = 200(14.28) + 0.5(20)(14.28)² = 4895.18 m

This is the distance the Mig-29 needs to travel in 14.28 seconds in order to avoid the missile. The assumption is that if it is slower than the missile, it will be hit. Let's find the acceleration of the Mig-29 required to miss the missile.

x_f,x = x_0,x + v_0,xt + 0.5a_xt²

4895.18 = 500 + 100(14.28) + 0.5a_x(14.28)²

a_x = 29.10 m/s²

(d) Can the Mig-29 escape disaster?

Let's define the Mig-29's maximum velocity in SI units first:

Mach 1.5 = 496.95 m/s

Now, let's see what the final velocity of the Mig-29 is when the acceleration in the x-direction is what is defined in problem (d).

v_f,x = v_0,x + a_xt

v_f,x = 100 + 29.1*14.28

v_f,x = 515.55 m/s

515.55 m/s > 496.55 m/s; Therefore, the Mig-29 cannot escape the missile.