91Ó°ÊÓ

Average mass of hail = 2g

speed of 15 m/s

Angle of fall = 45^o

Area of the window = 0.5 m² a

Rate of hail strike = 30 per second.

From the pressure force relationship we can calculate average force on window by

$\mathrm{Δ}{F}_x=\frac{\mathrm{Δ}{P}_x}{\mathrm{Δ}t}.....................................\left(1\right)$

Change in momentum can be defined as

ΔPx = mΔVx

So we get

role="math" localid="1650443757487" $\mathrm{Δ}{F}_x=\frac{m\mathrm{Δ}{v}_x}{\mathrm{Δ}t}..............................\left(2\right)$

We can calculate average pressure as

role="math" localid="1650443764918" $P=\frac{\mathrm{Δ}{F}_x}{A}.................................\left(3\right)$

As the hail strikes at an angle of 45^o. So we have to find the horizontal component as it will only add to pressure

$\mathrm{Δ}{v}_x=v\cos \mathrm{θ}-(-v\cos \mathrm{θ})=2v\cos \mathrm{θ}$

Substitute the given values, we get

$$\begin{gathered} \mathrm{Δ}{v}_x=2×(15m/s)×\cos 45 \\ \mathrm{Δ}{v}_x=(30m/s)×\frac{1}{\sqrt{2}} \\ \mathrm{Δ}{v}_x=21.21\mathrm{m}/\mathrm{s} \end{gathered}$$

Now calculate force by substituting values in equation (2)

$$\begin{gathered} \mathrm{Δ}{F}_x=\frac{(30×2×{10}^{-3}kg)×(21.21m/s)}{1s} \\ \mathrm{Δ}{F}_x=1.27\mathrm{N} \end{gathered}$$

Now calculate average pressure by substituting values in equation(3)

$$\begin{gathered} P=\frac{1.27N}{0.5m^2} \\ P=2.54{\mathrm{Nm}}^{-2} \end{gathered}$$

The value of atm pressure is 101325 N/m²

The pressure on the window is much less than that of atm. pressure.