91Ó°ÊÓ

The given data can be expressed below as:

• The initial height of the lander is 800 m.
• The value of c and d are $60.0\mathrm{m}/\mathrm{s}\mathrm{and}1.05\mathrm{m}/{\mathrm{s}}^2$.

This law illustrates that an object will continue to be in rest or in uniform motion unless it is resisted by an external force upon it.

Differentiating the equation of a motion along a straight line gives the initial velocity and using t =0, may be beneficial for the lunar lander.

The given equation can be expressed as:

$y\left(t\right)=800 m-(60.0 m/s)t+(1.05 m/s^2)t^2$

a) We know that,$v_y=\frac{dy}{dt}$

Hence, from the above equation, it can be expressed as:

$$\begin{gathered} v_y=\frac{d}{dt}\left(800-60t+1.05t^2\right) \\ =-60+2.1t \end{gathered}$$

The initial velocity of the lander at t =0 ,will be-

role="math" localid="1668400532951" $$\begin{gathered} v_{y0}=-60+2.1×0 \\ =-60\frac{m}{s} \end{gathered}$$

Thus, the initial velocity of the lander is$-60\frac{m}{s}$.

b) the lunar lander reaches the surface when$y\left(t\right)=0$

$1.05{\mathrm{t}}^2-60\mathrm{t}+800=0$

By solving the quadratic equation, we can get two roots which are ${\mathrm{t}}_1=35.9 \mathrm{s} \mathrm{and} {\mathrm{t}}_2=21.2 \mathrm{s}$. Hence, from the gathered roots, we will take$t_2=21.2 s$ as it is the least time to reach to the lunar surface.

The velocity of the lander before it reaches to the lunar surface is expressed as:

$$\begin{gathered} v_y=−60\frac{\mathrm{m}}{\mathrm{s}}+2â‹\dots 1\frac{\mathrm{m}}{{\mathrm{s}}^2}â‹\dots 21.2\mathrm{s} \\ =−15.5\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

Thus, the velocity of the lander just before it reaches to the lunar surface is $−15.5\frac{\mathrm{m}}{\mathrm{s}}$.