Projectile Motion- magnitude of final velocity

In summary, the problem involves a cannon on a 50m high cliff shooting a cannonball at an angle of 30° with an initial velocity of 200m/s. The cannonball lands 41m away from the cannon and the question asks for the magnitude of the velocity just before it hits the ground. Using projectile motion equations, the velocity components are found to be 173.2m/s and 97.68m/s. However, the given answer choices do not match the calculated velocity of 199m/s.
  • #1
Nelson2436
5
0

Homework Statement


Consider a cannon on a 50m high cliff. The cannon shoots a cannonball with an initial speed of 200m/s at an angle of 30°, causing the cannonball to land 41m away from the cannon. What is the magnitude of the velocity of the cannonball the instance just before it hits the ground?
a) -205m/s
b) 205m/s
c) 173m/s
d) 267m/s

Homework Equations


ax=0 ay=-g
1) Xf=Xi+Vixt
2) Yf=Yi+Viyt-½gt2
3) Vfy=Viy-gt
4) Vfy2=Viy2-2g(Yf-Yi)

The Attempt at a Solution


Vi=200m/s
Vix=200cos30 Viy=200sin30
Xi=0m Yi= 50m
Xf=41m Yf= 0m

In order to find the magnitude of Vf, I need to find its components, Vfx and Vfy. In projectile motion, Vfx=Vix=200cos30, so I just need to find Vfy.

Using equation 1 to find time: t=41/(200cos30)= 0.2367s
I then plugged time into equation 3 to find Vfy=200sin30-(9.8)(0.2367)= 97.68m/s

So Vfx=200cos30= 173.2m/s and Vfy= 97.68m/s so
Vf= √(Vfx2+Vfy2)= √(173.22+97.682)= 199 m/s

I don't see where I'm going wrong, but 199m/s is not one of the 4 multiple choice answers. I would greatly appreciate any help.
 
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  • #2
Hi, Nelson2436. Welcome to PF!

The 41 m given in the problem just doesn't seem right. You should actually be able to find where the cannon ball lands from the other information given in the problem. Did you state the problem exactly as given?
 
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  • #3
Yes, I stated the problem exactly as given. This problem is driving me crazy.
 

1. What factors affect the magnitude of the final velocity in projectile motion?

The magnitude of the final velocity in projectile motion is affected by three main factors: the initial velocity, the angle of launch, and the presence of air resistance. A higher initial velocity will result in a higher final velocity, while a lower initial velocity will result in a lower final velocity. The angle of launch also plays a role, with a higher angle resulting in a greater final velocity. The presence of air resistance can also decrease the magnitude of the final velocity.

2. How does the angle of launch affect the magnitude of the final velocity?

The angle of launch has a direct impact on the magnitude of the final velocity in projectile motion. The optimal angle for achieving the greatest final velocity is 45 degrees. This is because at this angle, the initial velocity is split equally between the horizontal and vertical components, resulting in the maximum vertical displacement and the greatest overall distance traveled.

3. Can the magnitude of the final velocity ever be greater than the initial velocity?

Yes, in certain cases the magnitude of the final velocity in projectile motion can be greater than the initial velocity. This can occur when the angle of launch is greater than 45 degrees, as the vertical component of the initial velocity will be greater and contribute more to the overall final velocity. However, this is only possible in the absence of air resistance.

4. How does air resistance affect the magnitude of the final velocity?

Air resistance plays a significant role in decreasing the magnitude of the final velocity in projectile motion. As an object moves through the air, it experiences air resistance which acts in the opposite direction of the motion. This resistance causes the object to slow down and results in a lower final velocity compared to what would be achieved in the absence of air resistance.

5. Is there a mathematical equation for calculating the magnitude of the final velocity in projectile motion?

Yes, there is a mathematical equation for calculating the magnitude of the final velocity in projectile motion. It is given by the formula Vf = √(Vi² + 2gh), where Vf is the final velocity, Vi is the initial velocity, g is the acceleration due to gravity, and h is the vertical displacement. This equation assumes no air resistance and a launch angle of 45 degrees.

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