# Physics Projectile Question

1. Oct 8, 2009

### Leo34005

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

2. Relevant equations

Range = v0^2 sin(2 theta)/g
vy(tO=v0sin(theta) - gt

3. The attempt at a solution

a) the range equation is

Range = v0^2 sin(2 theta)/g
Range =1590^2sin(94)/9.8=247,341 m

there are several ways to find the time in the air...the vertical velocity varies as:

vy(tO=v0sin(theta) - gt

this will be zero when t=v0sin(theta)/g =
1590sin47/9.8=118.7s

the vertical velocity is zero at the apex of motion, which occurs at the midway point of the flight, therefore the total time in the air is twice this or

237.3 s

b) we use the range equation again, but this time to find v0:

range = v0^2sin(2 theta)/g

v0=sqrt[g R/sin(2 theta)]=
sqrt[9.8x9.9/sin(46)]=11.6m/s

neglecting air friction, the speed on landing will be equal to the speed at launch

I think i got something wrong not sure though

2. Oct 8, 2009

### kuruman

Looks OK. The given speed of the shell looks rather high but that's not your doing.