Question

PLS HELP DUE IN 3 HOURS!

Answer 1

Judging by the sketch, H is the vertical distance and D is the horizontal distance between where the bullet is shot and where it gets lodged in the wall. Let v₀ = 374 m/s, h = 0.45 m, and d = 9.6 m.

At time t, the bullet has

• horizontal position

$x = v_0 t$

• vertical position

$y = H - \dfrac12 gt^2$

Let t₁ and t₂ be the times when the bullet reaches the window and the wall, respectively. We want to find H and D given that

$\begin{cases}D = v_0 t_1 \\ h = H - \dfrac12 g {t_1}^2 \\ D + d = v_0 t_2 \\ 0 = H - \dfrac12 g {t_2}^2\end{cases}$

Eliminate D and H to get a system of equations involving t₁ and t₂ :

$(D + d) - D = v_0 t_2 - v_0 t_1 \implies -t_1 + t_2 = \dfrac d{v_0}$

$\left(H - \dfrac12 g {t_1}^2\right) - \left(H - \dfrac12 g {t_2}^2\right) = h - 0 \implies -{t_1}^2 + {t_2}^2 = \dfrac{2h}g$

The first of these equations says

$t_2 = \dfrac d{v_0} + t_1$

and substituting into the second equation gives

$-{t_1}^2 + \left(\dfrac d{v_0} + t_1\right)^2 = \dfrac{2h}g$

Solve for t₁ :

$\dfrac{d^2}{{v_0}^2} + \dfrac{2d}{v_0} t_1 = \dfrac{2h}g$

$\implies t_1 = \dfrac{hv_0}{gd} - \dfrac{d}{2v_0}$

Solve for t₂ :

$\implies t_2 = \dfrac{hv_0}{gd} + \dfrac{d}{2v_0}$

Now solve for H and D :

$H = \dfrac12 g {t_2}^2$

$H = \dfrac g2 \left(\dfrac{hv_0}{gd} + \dfrac{d}{2v_0}\right)^2$

$H = \dfrac{h^2{v_0}^2}{2gd^2} + \dfrac h2 + \dfrac{gd^2}{8{v_0}^2}$

$H = \dfrac{(0.45\,\mathrm m)^2 \left(374\frac{\rm m}{\rm s}\right)^2}{2g(9.6\,\mathrm m)^2} + \dfrac{0.45 \,\rm m}2 + \dfrac{g(9.6\,\mathrm m)^2}{8\left(374\frac{\rm m}{\rm s}\right)^2} \approx \boxed{16 \mathrm m}$

$D = v_0 t_1$

$D = v_0 \left(\dfrac{hv_0}{gd} - \dfrac{d}{2v_0}\right)$

$D = \dfrac{h{v_0}^2}{gd} - \dfrac d2$

$D = \dfrac{(0.45\,\mathrm m) \left(374\frac{\rm m}{\rm s}\right)^2}{g(9.6\,\mathrm m)} - \dfrac{9.6}2 \,\mathrm m \approx \boxed{660 \,\mathrm m}$