4. Refer to the diagram below. A truck-mounted cannon must send a shell to a tar

Question

4. Refer to the diagram below. A truck-mounted cannon must send a shell to a target located at point B on a slope Unfortunately, the cannon angle, ' is fixed with respect to the level truck bed (the elevation pivot is stuck, apparently) so that firing the cannon from any stationary position on the available roadway would overshoot the target. (The speed, vat which a shell leaves the gun's muzzle-measured with respect to the gun-is too fast.) A decision is made to fire the cannon while the truck is moving -driving at constant speed to the left on the level road The shot is fired at the moment when the cannon's muzzle passes directly over the firing point A, as shown. The shell strikes the target after spending a total time as a projectile. (slope) (road) What is the reading on the truck's speedometer when the shot is fired? Here is a summary of known values: ,.02, Vmutzle, D, R, , g

Explanation / Answer

4. given truck mounted canon, target at point B
canon angle = theta ( fixed)
muzzle velocity vm = constant
velocity of truck to the left = v
shell fired at position A
time of flight = dT

so, horizontasl velocity of projectile = vmcos(theta) - v
hence (vmcos(theta) - v)*dT = D + Rcos(theta2)

hence
v = vmcos(theta) - (D + Rcos(theta2))/dT

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