Mr. Smith is flying his single-engine plane at an altitude of 2400 feet. He sees a cornfield at an angle of depression of 30º. What is his horizontal distance to the cornfield?

Height of the fighter plane =1.5km=1500 m

Speed of the fighter plane, v=720km/h=200 m/s

Let be the angle with the vertical so that the shell hits the plane. The situation is shown in the given figure.

Muzzle velocity of the gun, u=600 m/s

Time taken by the shell to hit the plane =t

Horizontal distance travelled by the shell =u

x

t

Distance travelled by the plane =vt

The shell hits the plane. Hence, these two distances must be equal.

u

x

t=vt

u Sin θ=v

Sin θ=v/u

=200/600=1/3=0.33

θ=Sin

−1

(0.33)=19.50

In order to avoid being hit by the shell, the pilot must fly the plane at an altitude (H) higher than the maximum height achieved by the shell for any angle of launch.

H

max

=u

2

sin

2

(90−θ)/2g=600

2

/(2×10)=16km

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2