Calculating the pressure of a hailstone

[pentazoid]

Homework Statement

During a hailstorm , hailstones with an average mass of 2 g and a speed 15m/s strike a window pane at a 45 degree angle. The area of the window is .5 m^2 and the hailstones hit it at a rate of 30 per second. What average presure do they exert on the window? How does this compare to the pressure of the atmosphere?

Homework Equations

$$\overline{}$$P=$$\overline{}$$F/A=-m($$\Delta$$v_x/($$\Delta$$t)/A

The Attempt at a Solution

N/s=30 hailstones/sec
m=2 grams
$$\Delta$$v_x=v_x,final-v_x,initial=(-v_x-v_x)=-2v_x

P=m(-2v_x)/($$\Delta$$t))*N/A=(2e-3 kg)(-2(15)(cos 45 degrees)(30))/(second)/(.5 m^2)

to compare the pressure of the hail stone to the pressure of the atmosphere, wouldn't I take the ratio of the pressure of the hailstone to the pressure of the atmosphere?

 

[Dick]

The pressure calculation looks ok to me. Sure, a ratio would be a good way to compare the two pressures.