91Ó°ÊÓ

The initial velocity is in x direction, thus it is

$$\begin{gathered} v_{0x}=250m/s \\ h=45m \end{gathered}$$

This problem is based on kinematic equations that describe the motion of an object with constant acceleration. By applying second kinematic equation, time of the projectile can be found. Further, using the velocity of the projectile, the horizontal distance can be calculated. And the magnitude of the vertical velocity of the projectile as it strikes the ground can be computed using first kinematic equation.

Formula:

The displacement in kinematic equation is given by,

$d=v_{0}t+\frac{1}{2}a{t}^2$ (i)

The velocity in general is,

$v_0=\frac{d}{t}$ (ii)

$v=v_0+at$ (iii)

It is given that,

$$\begin{gathered} \mathrm{d}-{\mathrm{v}}_0\mathrm{t}+\frac{1}{2}{\mathrm{at}}^2 \\ \mathrm{a}=\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{As}\mathrm{h}=45.0\mathrm{m} \\ {\mathrm{v}}_{0\mathrm{y}}=0\mathrm{m}/\mathrm{s} \\ \mathrm{t}=\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}} \\ \mathrm{t}=\sqrt{\frac{2\left(45.0\mathrm{m}\right)}{9.8\mathrm{m}/{\mathrm{s}}^2}} \\ \mathrm{t}=3.03\mathrm{s} \end{gathered}$$

It is given that,

$$\begin{gathered} {\mathrm{v}}_0=\frac{\mathrm{d}}{\mathrm{t}} \\ \mathrm{As}{\mathrm{v}}_{0\mathrm{x}}=250\mathrm{m}/\mathrm{s} \\ \mathrm{t}=3.03\mathrm{s} \\ {\mathrm{d}}_{\mathrm{x}}=\left(250\frac{\mathrm{m}}{\mathrm{s}}\right)\left(3.03\mathrm{s}\right) \\ {\mathrm{d}}_{\mathrm{x}}=758\mathrm{m} \end{gathered}$$

It is given that,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{y}}={\mathrm{v}}_{0\mathrm{y}}+\mathrm{gt} \\ {\mathrm{v}}_{0\mathrm{y}}=0\mathrm{m}/\mathrm{s} \\ \mathrm{As},\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{t}=3.03\mathrm{s} \\ {\mathrm{v}}_{\mathrm{y}}=\left(9.8\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\left(3.03\right) \\ {\mathrm{v}}_{\mathrm{y}}=29.7\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, using the concept of free fall and kinematic equations, the time of flight and range of the projectile with given conditions can be found.