what is the required velocity of a cannonball fire at 30 degrees to hit a target 240 feet away?

Question

what is the required velocity of a cannonball fire at 30 degrees to hit a target 240 feet away

Answer 1

i assume the target is on the same horizontal level as the cannon, and that the cannon will follow a perfect parabolic trajectory that ignores air resistance.

i will be resolving the angled velocity into its vertical and horizontal components. the logic is that the time it takes for the cannon to reach its peak and come back down is the same time it takes for the cannon to reach the target horizontally.

horizontal velocity component, Vcos30. time required to reach target, t1= 240/Vcos30.

vertical velocity component, Vsin30. vertical velocity at the peak of the trajectory is 0 due to gravity causing a change in direction. the cannon at this point starts to fall and takes the same amount of time to go down as it took to go up. time taken to go reach the peak, t2=(Vsin30)/g, where g is 32.185.

for the cannon to reach the target exactly when it reaches down to the bottom of the trajectory, t1=2x t2.

240/Vcos30=2(Vsin30)/32.185

V²(cos30)(sin30)=240 x 32.185 / 2
V=sqrt(8919)
V=94.44 feet per second

Answer 2

240/Vcos30=2(Vsin30)/32.185 V²(cos30)(sin30)=240 x 32.185 / 2 V=sqrt(8919) V=ninety 4.40 4 ft in keeping with 2d you could't answer this questions without the load of the cannonball and the diameter of the Ball it self, Reversing the equation dose no longer provide me the load

Answer 3

If I was storming the castle and I had cannon this would be a very important question.

As an excercise in pure physics it could be interesting if I had more info and a reason.

Goodnight,
Jonnie

Answer 4

Vx = Vcos30
Vy = Vsin30

0 - Vy = -a(t-0)
t = Vy/a

240 = 2Vcos30t = 2V^2cos30sin30)/a
V^2 = 240a/(2*0.5*(√3/2) =
V^2 = 480(32.2)/√3
V = 94.464 ft/sec^2

Answer 5

ok i don't know physics, but i assume amount of powder, how long the cannon would be, the weight and the fitting of the projectile./ whether of not it is humid, dry, to many variables, may i suggest trial and error, start with a small charge and go from there.

Answer 6

210 mph

Answer 7

you probably would have to know how much the cannon ball weighs too. and wind speed maybe