Officer R. Ingling, I appreciate you trusting my calculus professor’s recommendation. Hopefully I won’t disappoint. Let’s go over the issue at hand again. Bobo the clown was shot out of a cannon, across a canyon, and into the canyon wall, killing him. You have a suspect in hand, the owner of the traveling circus in which Bobo worked, Mr. Rasterdly, who was near when Bobo’s cannon was shot and who you assume to have been the one to press the button to shoot the cannon. However you would like to be sure that Mr. Rasterdly shot the cannon and that it wasn’t a potential suicide before you lock him away. Let’s also go over some of the technical information you’ve given me. One end of Bobo’s cannon is three feet off the ground and the whole cannon …show more content…
Because 1 meter = 3.28 feet, we’ll just divide the feet by 3.28 to get our new measurements. 5ft3.28ft = 1.524 meters 3ft3.28ft = 0.915 meters A moving object’s position is given by the vector-valued function r(t) = for t0, the object’s velocity is the derivative of the position which is given by v(t) = r’(t) = , and the speed of the object is given by |v(t)| = x’(t)2+ y’(t)2(Briggs, 834). To find x(t) and y(t) we’re going to use the information we have on the cannon. Let’s visualize it as a triangle, like so: We’re looking for the angle marked so to find it we use the formula for sin(x) which tells us to take the opposite value (0.915m) over the hypotenuse value (1.524m) and plug it into a formula using arcsin, which will give us the angle. So we have: sin-1(x) = 0.9151.524= 36.9 Now to find the horizontal and vertical vectors, or x(t) and y(t), we’re going to use the speed at which the cannon fired Bobo (30mps) and the angle we just found. For x(t) we use cosine because it’s going in a horizontal direction and for y(t) we use sine because it’s going in a vertical direction. Our formulas are as follows: x(t) = 30cos36.9 = 24 y(t) = 30sin36.9 = 18 Next to find
5.95ohms, 105degrees-9.76ohms, 120degrees-5.24ohms, 135degrees-9.20ohms, 150degrees-12.80ohms, 165degrees-16.58ohms, 180degrees-5.24ohms. These results varied, it appears as though when I made a slight increment to the degrees it made a drastic effect to the meter
We want to minimize this function. Using calculus we find the minimum to be at t approximately 1.15 seconds and velocity approximately 36.66 feet per
The sketch below shows the trajectory of a projectile propelled with an initial velocity V0 at an angle 8. Neglecting air resistance the horizontal range travelled is given
Let the mass of the fired bullet be Mb. It is fired at an initial velocity of v. The momentum
(Melvin, Mangonel - “Physics of Catapults”) The speed and distance of the projectile depended on how much force the catapult applied to the projectile, and the momentum depended on the mass and the velocity of the projectile (dead diseased cow, or flaming
1. Move the slider all the way to accurate, click on the tape measure and the grid.
Using this relationship assumes that the force constant is constant, or that moving the arms back 2 meters gives twice the force that moving them back 1 meter would do, which is most likely not correct, but close enough for a general assumption of the force to be made. When the force is applied, the projectile is accelerated to the end of the ballista, at which point it released with a velocity v and an angle q from the horizontal. The velocity can be found using the kinematic equation
Chapter 3 Falling Objects and Projectile Motion Gravity influences motion in a particular way. How does a dropped object behave? ! Does the object accelerate, or is the speed constant? !
The physics behind a catapult can be a complex topic to grasp. But, when broken down, looking individually at their historical uses, design types, and related physics concepts, it’s surprisingly simple to form an opinion on which catapult design is most effective and why. These categories, when combined, define the uses, functionality, and effectiveness of the historical (and modern day) catapult.
This catapult experiment requires a rubber band to be stretched by an individual while another individual presses a matchbox on it, pulling it back and then letting go to allow to matchbox to be flung into the air. From that, we will then change the variable, which will be the thickness of the rubber band to differ the distance the matchbox travels. This will happen due to potential and kinetic energy.
The velocity, or speed, in a cannonball during the 19th century varied greatly. It all depended on the angle degree the cannon was facing, the size of the cannonball, and many other things. Air resistance and gravity also played
The cannon arm is designed to be 4” in diameter and will be displaced by a total of 0.964” when every stage is complete and the clutch is fully locked. By completing each cannon displacement in 0.0117 seconds or less, the top speed of the cannon arm is 39.6 ft/s. The maximum air pressure necessary to hold the cannon in place is 15.87 psig, which will necessitate a max oil pressure of 79.33
In this experiment, a photogate, a chopper, and a Universal Lab Interface were used to determine the free fall motion of the chopper as it was released. A ball, carbon paper, and an L-shape projector were also used to determine the range of projectile motion of a ball being released from a horizontal yet slightly vertical slope. At the end of the experiment, one will know how velocity
A shell fired in a gun accelerates in the barrel over a length of 1.5 m to the exit