# Homework Help: Projectile Motion Cannon Ball Question

1. Oct 4, 2012

### tatterspwn

1. The problem statement, all variables and given/known data
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°

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

3. 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?

2. Oct 4, 2012

### LawrenceC

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. Oct 4, 2012

### tiny-tim

hi tatterspwn!
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

4. Oct 4, 2012

### tatterspwn

For that I did
Tt=2 (Vsinθ)/g
Tt=2 (6.552sin35)/9.8
Tt= 0.7670s

5. Oct 4, 2012

### tiny-tim

ah now i see, you did …
… yes, that seems fine

6. Oct 4, 2012

Thank you

7. Oct 5, 2012

### LawrenceC

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.