A cannonball has just been shot out of a cannon aimed 45∘ above the horizontal rightward direction. Drag forces cannot be neglec
1 answer:
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.
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1) 5.79 s
2) 98.4 ft/s
Explanation:
1)
The motion of the car is a uniformly accelerated motion (it means it travels with constant acceleration), so we can find the time it takes for the car to stop by using the following suvat equation:
where
s is the distance travelled
v is the final velocity
t is the time
a is the acceleration of the car
In this problem we have:
s = 285 ft is the distance travelled
is the acceleration of the car (negative since the car is slowing down)
v = 0 ft/s is the final velocity of the car, since it comes to a stop
Solving for t, we find:
2)
The initial speed of the car can be found by using another suvat equation, namely:
where
v is the final speed
u is the initial speed
a is the acceleration
t is the time
In this problem, we have:
v = 0 is the final speed of the car
is the acceleration of the car (negative since the car is slowing down)
t = 5.79 s is the total time of motion (found in part 1)
Therefore, the initial speed of the car is: