Answer:
The free-body diagram of the cannonball is found in the attachment below
<em>Note The question is incomplete. The complete question is as follows:</em>
<em>A cannonball has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction. Drag forces cannot be neglected.</em>
<em>Draw the free-body diagram of the cannonball.</em>
Explanation:
Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.
In order to construct free-body diagrams, it is important to know the various types of forces acting on the object in that situation. Then, the direction in which each of the forces is acting is determined. Finally the given object is drawn using any given representation, usually a box, and the direction of action of the forces are represented using arrows.
In the given situation of a cannonball which has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction., the forces acting on it are:
F = force exerted by the cannon acting in the direction of angle of projection
Fdrag = drag force. The drag force acts in a direction opposite to the force exerted by the cannon
Fw = weight of the cannonball acting in a downward direction
The free body diagram is as shown in the attachment below.
Explanation:
It is given that,
Radius of the front sprockets,
Radius of the rear sprockets,
The angular speed of the front sprocket is 12.3 rad/s,
(a) Linear speed of the front sprockets,
Linear speed of the rear sprockets,
(b) Let is the centripetal acceleration of the chain as it passes around the rear sprocket.
`Hence, this is the required solution.
Answer:
a) Maximum height = 36.6 m
b) Horizontal distance at which the ball lands = 166.1 m
c) x-component = 32 m/s. y-component = - 27 m/s
Explanation:
Please, see the attached figure for a description of the problem.
The velocity vector "v" of the cannonball has two components, a horizontal component, "vx", and a vertical component "vy". Notice that at the maximum height, the vertical component "vy" of the velocity vector is 0.
In the same way, the position vector "r" is composed by "rx", its horizontal component, and "ry", the vertical component.
The velocity vector "v" ad the position vector "r" at time "t" are given by the following equations:
v = (v0 * cos α, v0 * sin α + g * t)
r = (x0 + v0 * t * cos α, y0 + v0 * t * sin α + 1/2 * g * t²)
Where
v0 = magnitude of the initial velocity vector
α = launching angle
g = gravity acceleration (-9.8 m/s², because the y-axis points up)
t = time
x0 = initial horizontal position
y0 = initial vertical position
If we consider the origin of the system of reference as the point at which the cannonball leaves tha catapult, then, x0 and y0 = 0
a) We know that at maximum height, the vertical component of the vector "v" is 0, because the ball does not move up nor down at that moment (see figure). Then:
0 = v0 * sin α + g * t
-v0 * sin α / g = t
-40 m/s * sin 37° / -9.8 m/s² = t
t = 2.5 s
We can now calculate the position of the cannonball at time t=2.5 s to obtain the maximum height:
r = (x0 + v0 * t cos α, y0 + v0 * t * sin α + 1/2 * g * t²)
The max height is the magnitude of the vector ry max (see figure). The vector ry max is:
ry = (0, y0 + v0 t sin α + 1/2 g * t²)
magnitude of ry = )
Then:
max height = y0 + v0 * t * sin α + 1/2 * g * t²
max height = 0 m + 40 m/s * 2.5 s * sin 37° - 1/2* 9.8 m/s² * (2.5 s)² = 29.6 m
Since the ball leaves the catapult 7 m above the ground, the max height above the ground will be 29.6 m + 7 m = 36.6m
<u>max height = 36.6 m</u>
b) When the ball hits the ground, the position is given by the vector "r final" (see figure). The magnitude of "rx", the horizontal component of "r final", is the horizontal distance between the catapult and the wall.
r final = ( x0 + v0 * t * cos α, y0 + v0 * t * sin α + 1/2 * g * t²)
We know that the vertical component of "r final" is -7 (see figure).
Then, we can obtain the time when the the ball hits the ground:
y0 + v0 * t * sin α + 1/2 * g * t² = -7 m
0 m + 40 m/s * t * sin 37° + 1/2 g * t² = -7 m
7 m + 40 m/s * t * sin 37° + 1/2 (-9.8 m/s²) * t² = 0
7 m + 24.1 m/s * t - 4.9 m/s² * t² = 0
solving the quadratic equation:
t = 5.2 s (The negative solution is discarded).
With this time, we can calculate the value of the horizontal component of "r final"
Distance to the wall = |rx| = x0 + v0 t cos α
|rx| = 0m + 40 m/s * 5.2 s * cos 37° =<u> 166.1 m</u>
c) With the final time obtained in b) we can calculate the velocity of the ball:
v = (v0 * cos α, v0 * sin α + g * t)
v =(40 m/s * cos 37°, 40 m/s * sin 37° -9.8 m/s² * 5.2 s)
v =(32 m/s, -27 m)
x-component = 32 m/s
y-component = - 27 m/s
