- #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

Vf

^{2}-Vi

^{2}/2g

Vx=Vsinθ

Vy=Vcosθ

Tt=2(Vsinθ)/g

Vy=Vi+at

Dx=Vxt

Dy=Vit+at

^{2}/2

## The Attempt at a Solution

2.19= 0

^{2}-vi

^{2}/-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+at

^{2}/2

d=3.758(0.3835)+(-9.8x0.3835

^{2})/2

d=0.7205

I concluded with:

Tt= 0.7670s

Max height= 0.7205m

Range= 2.882m

Is this correct?