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

Question

"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 "

Where did the 32.185 come in? How do you get it?

Answer 1

Your velocity is in ft/s so g should be in feet/s^2 to be consistent. To answer your question directly: 32.185 must have come from g=9.81*3.28=32.17. Not exactly the same but close. There's no other explanation, tho, as to where 32.185 came in. Note that in the MKS (MeterKilogramSecond) system of units g=9.81m/s^2. To convert meters to feet, I used the equivalent of 3.28ft=1m.

If you want to do this problem another way convert 240 ft to meters by dividing it by 3.28. Then use g=9.81m/s^2. The velocity you get will then be in m/s.

Answer 2

240/Vcos30=2(Vsin30)/32.185
V²(cos30)(sin30)=240 x 32.185 / 2
V=sqrt(8919) V=94.44 feet per second
You can't answer this questions with out the weight of the cannonball and the diameter of the Ball it self, Reversing the equation dose not give me the weight

Answer 3

Won't you need the weight or mass of the cannonball?