Answer:
<em>The horizontal velocity vector of the canonball does not change at all, but is constant throughout the flight.</em>
Explanation:
First, I'll assume this is a projectile simulation, since no simulation is shown here. That been the case, in a projectile flight, there is only a vertical component force (gravity) acting on the body, and no horizontal component force on the body. The effect of this on the canonball is that the vertical velocity component on the canonball goes from maximum to zero at a deceleration of 9.81 m/s^2, in the first half of the flight. And then zero to maximum at an acceleration of 9.81 m/s^2 for the second half of the flight before hitting the ground. <em>Since there is no force acting on the horizontal velocity vector of the canonball, there will be no acceleration or deceleration of the horizontal velocity component of the canonball. This means that the horizontal velocity component of the canonball is constant throughout the flight</em>
There is no acceleration in the horizontal direction because the speed in that direction is constant. So it stays at zero.
Answer:
) the uniform disk has a lower moment of inertia and arrives first.
Explanation:
(a) the uniform disk has a lower moment of inertia and arrives first.
(b) Let's say the disk has mass m and radius r, and
the hoop has mass M and radius R.
disk: initial E = PE = mgh
I = ½mr², so KE = ½mv² + ½Iω² = ½mv² + ½(½mr²)(v/r)² = (3/4)mv² = mgh
m cancels, leaving v² = 4gh / 3
hoop: initial E = Mgh
I = MR², so KE = ½MV² + ½(MR²)(V/R)² = MV² = Mgh
M cancels, leaving V² = gh
Vdisk = √(4gh/3) > Vhoop = √(gh)
