10. A military jet flies horizontally with a velocity of 450 m/s and 20,000 m above the ground. When the jet is straight above an artillery gun a shell is fired. Assuming the shell hits the jet.

Question

a. Calculate the horizontal component of the initial velocity of the shell.
b. Calculate the vertical component of the initial velocity of the shell.

Answer 1

Not enough data. The shell could have been fired fast and steep, or slow and shallow, and could still have hit the plane.

If the shell is fired with speed v and at angle θ, its horizontal speed is v cosθ. So, if it hits the plane at time t, we have

v cosθ t = 450t
cosθ = 450/v

Now, the height y is

v sinθ t - 4.9t^2, or
v√(1-(450/v)^2) t - 4.9t^2

So, we need

v√(1-(450/v)^2) t - 4.9t^2 = 20000

So, the shell could be fired at 10000m/s and hit the plane in 2 seconds.
Or, it could be fired at 1000m/s and hit after 26 seconds.

What we can say is that the initial speed must be at least 770 m/s, or it won't fly high enough to reach 20000m before it starts to fall back to earth.