Answer to Question #169145 in Physics for Rijak

Question #169145

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(𝑑) = 800 π‘š βˆ’ (60.0 π‘š/𝑠) 𝑑 + (1.05 π‘š/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?


1
Expert's answer
2021-03-08T08:27:21-0500

(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}."

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