Answer to Question #169145 in Physics for Rijak

Question

A lunar lander is descending toward the moon’s surface. Until the lander reaches the

surface, its height above the surface of the moon is given by 𝑦(𝑡) = 800 𝑚 − (60.0 𝑚/𝑠) 𝑡 + (1.05 𝑚/s²) t² .

(a) What is the initial velocity of the lander, at t=0?

(b) What is the velocity of thel landerjust before it reaches the lunar surface?

Answer 1

(a) Velocity is the derivative of position with respect to time:

$v_y=\dfrac{dy}{dt},$ $v_y=\dfrac{d}{dt}(800-60t+1.05t^2)=-60+2.1t.$

Finally, we can find the initial velocity of the lander, at $t=0$ :

$v_{y0}=-60+2.1\cdot0=-60\ \dfrac{m}{s}.$

(b) The lunar lander reaches the lunar surface when $y(t)=0$:

$1.05t^2-60t+800=0$

This quadratic equation has two roots: $t_1=35.9\ s$ and $t_2=21.2\ s.$ The correct answer is $t=21.2\ s$ (we choose the least time to land to the lunar surface).

Finally, we can find the velocity of the lander just before it reaches the lunar surface:

$v_y=-60\ \dfrac{m}{s}+2.1\ \dfrac{m}{s^2}\cdot21.2\ s=-15.5\ \dfrac{m}{s}.$