Answer:
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
To find the answer, you need to solve for b.
2(2b + 3) = 12 divide both sides by 3
2b + 3 = 6 divide both sides by 2
2b = 3 subtract 3 from both sides
b = 1.5 divide both sides by 2
b = 1.5 satisfies the equation. if you plug 1.5 in, you get that the equation is true.
(y2-y1)÷(x2-x1) so 11.5-2.5 divided by -2.6+5.6 and the answer is 3
The answer is y=4x+3, it's already in slope intercept form.
