A Canon shoots an artillery shell towards a target 4.54 km distant, where it lands at the same level it was shot. It was noted that the elapsed time of the projectile was 27.5 seconds. What was the muzzle velocity of the shell?

Question

(there is a picture showing a canon shooting the muzzle over a mountain, but the angle was not given)

Answer 1

you know horizonal velocity (distance/time).

from time in air:
in the vertical component...
Vf=vi-4.9t^2
-vi*sinTheta=vi*sinTheta-4.9t^2, you know time inair, so solve for vi*sinTheta
a) vi*sinTheta=4.9*27.5^2/2

from horizonal velocity
b) vi*cosTheta=4.54/27.5

Now, divide a) by b) and you get an expression for
tanTheta= 4.9*27.5^2/2 / 4.54/27.5
and you can solve for Theta. From that, get Vi