Below you'll find a video showing a tank doing some target practice. It's awesome because you get to see it in a bunch of different angles and in slow motion!
Alrighty, all you aspiring engineers, this one's for you (although if a Chemistry major or NROTC student could solve this I'd be really impressed!).
So, let's say you knew that the tank stood 500 meters from the target. The tank and the target are on the same ground level vertically, but the barrel of the gun on the tank is 2 meters off the ground. It takes 3 seconds total for the target to be hit. Suppose that the barrel of the gun was angled at 30 degrees (theta = 30) with respect to the (+) x-axis.
WHAT IS THE INITIAL VELOCITY OF THE PROJECTILE?
By the way, this is a classic problem you would find in Dr. Sayar's Dynamics class!
BONUS:
YOU'LL NOTE THAT THE ABOVE PROBLEM WAS TOO EASY BECAUSE I GAVE YOU SO MUCH INFORMATION. NOW, I'LL AWARD A CANDY BAR OF YOUR CHOICE TO THE PERSON WHO CAN FIND INITIAL VELOCITY WITHOUT KNOWING THE ANGLE OF THE BARREL (THETA = ???) . YOU'LL HAVE TO USE 2 EQUATIONS TO FIND THE TWO UNKNOWNS. ALL YOU ENGINEERS SHOULD DO THIS SINCE YOU'LL HAVE TO DO IT EVENTUALLY IN DYNAMICS!!!
You can use the range equation to solve this since you have theta and the distance. The initial velocity should be approximately 127 meters per second. However this answer is if it started at y=0.
ReplyDeleteInteresting video
We just started learning about force this week, so my question to you: Is there a way to calculate the net force that is exerted on the tank when it fires the projectile in Newtons?
ReplyDeleteElamon White
Mark, I calculated initial velocity at 88.2 meters per second when y=0. In this case "y" refers to the height of the barrel of the tank, right? The formula I used was: y=y0+V0*sin(30)-(1/2)*g*t^2. What formula did you use?
ReplyDeleteHas anybody else tried this problem? What answer did you get?
Elamon, yes, there most certainly is a way to calculate the net force. In the case of this video you saw with the tank it is quite easy. You'll notice that the tank stayed stationary on the ground even after the explosion. Since the tank did not move left, right, up or down, then its acceleration was 0 and so according to Newtons second law, Fnet=ma, with a=0, then Fnet=0 also.
ReplyDelete