Projectile Motion Cannon Ball Question

In summary: So how did you get those answers? In summary, the problem involves finding the initial velocity, time of flight, horizontal and vertical velocities, and horizontal and vertical displacement of a projectile launched at a 35 degree angle from a cannon that previously launched a ball at 90 degrees and reached a height of 2.19m. Using the appropriate equations, the initial velocity was found to be 6.552 m/s and the time of flight to be 0.7670 seconds. The horizontal velocity was calculated to be 3.758 m/s and the vertical velocity to be 5.367 m/s. The horizontal displacement was determined to be 2.882 m and the vertical displacement at 1/2 time to be
  • #1
tatterspwn
10
0

Homework Statement


A cannon shot a ball at 90° and reached 2.19m high. From this I need to find Vi of that and find time, Vx, Vy, Dx, and Dy of a projectile being shot from this cannon at 35.00°.
I would just like to confirm I have done this correctly.
Dy=2.19m
θ=90.00°

Homework Equations


Vf2-Vi2/2g
Vx=Vsinθ
Vy=Vcosθ
Tt=2(Vsinθ)/g
Vy=Vi+at
Dx=Vxt
Dy=Vit+at2/2

The Attempt at a Solution


2.19= 02-vi2/-19.6
Vi=6.552m/s

Tt=2(6.552sin35)/9.8
Tt= 0.7670s

Vx=3.758m/s
Vy= 5.367m/s

Dx= Vxt
Dx= 3.758(0.7670)
Dx=2.882m

To find height at 1/2 time:
d=Vit+at2/2
d=3.758(0.3835)+(-9.8x0.38352)/2
d=0.7205

I concluded with:
Tt= 0.7670s
Max height= 0.7205m
Range= 2.882m

Is this correct?
 
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  • #2
For something shot upward at an angle of 35 degrees how can the following be true? Vy > Vx ?

Vx=3.758m/s
Vy= 5.367m/s
 
  • #3
hi tatterspwn! :smile:
tatterspwn said:
A cannon shot a ball at 90° and reached 2.19m high. From this I need to find Vi of that and find time, Vx, Vy, Dx, and Dy of a projectile being shot from this cannon at 35.00°.

your vi for 90° looks ok,

so your vix and viy for 35° should also be ok

but you need to find t for 35° before going any further :wink:​
 
  • #4
tiny-tim said:
hi tatterspwn! :smile:


your vi for 90° looks ok,

so your vix and viy for 35° should also be ok

but you need to find t for 35° before going any further :wink:​

For that I did
Tt=2 (Vsinθ)/g
Tt=2 (6.552sin35)/9.8
Tt= 0.7670s
 
  • #5
ah now i see, you did …
tatterspwn said:
I concluded with:
Tt= 0.7670s
Max height= 0.7205m
Range= 2.882m

Is this correct?

… yes, that seems fine :smile:
 
  • #6
tiny-tim said:
ah now i see, you did …


… yes, that seems fine :smile:

Thank you :approve:
 
  • #7
There remains a problem here. The initial velocity of 6.55 m/s is correct. The time of flight is correct also. However you are mixing up the Vx and Vy velocities. So again I ask how a projectile shot at an angle of 35 degrees from horizontal can have a vertical velocity greater than its horizontal velocity? You have written:

Vx=3.758m/s
Vy= 5.367m/s

Vy cannot be greater than Vx for an angle of 35 degrees. Sin 35 < cos 35.
 

FAQ: Projectile Motion Cannon Ball Question

1. How does air resistance affect the motion of a cannon ball?

As a scientist, I can tell you that air resistance can have a significant impact on the motion of a cannon ball. As the cannon ball travels through the air, it experiences a force in the opposite direction of its motion due to air resistance. This force can slow down the cannon ball and alter its trajectory.

2. What factors affect the range of a cannon ball?

The range of a cannon ball is affected by several factors, including the initial velocity of the cannon ball, the angle at which it is fired, and the presence of air resistance. The range can also be affected by the mass and shape of the cannon ball, as well as any external forces acting on it.

3. How do you calculate the velocity of a cannon ball at a given time during its flight?

To calculate the velocity of a cannon ball at a given time, you can use the equations of motion for projectile motion. These equations take into account the initial velocity, acceleration due to gravity, and time elapsed since the cannon ball was fired. By plugging in the appropriate values, you can determine the velocity at any point during the cannon ball's flight.

4. Can a cannon ball ever reach a height of zero during its flight?

Yes, a cannon ball can reach a height of zero during its flight. This would occur at the highest point of the cannon ball's trajectory, also known as the apex. At this point, the vertical velocity of the cannon ball would be zero, but it would still have a horizontal velocity due to its initial velocity.

5. How does the angle of elevation affect the maximum height and range of a cannon ball?

The angle of elevation, or the angle at which the cannon ball is fired, can greatly impact both the maximum height and range of the cannon ball. As the angle increases, the maximum height of the cannon ball also increases. However, there is an optimal angle that will result in the maximum range for a given initial velocity and air resistance. This angle is typically around 45 degrees.

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