Solve the problem.
-If a rock is thrown vertically upward from the surface of the moon at a speed of $21 \mathrm{~m} / \mathrm{s}$, its height after $t$ seconds will be $s(t)=21 t-0.8 t^{2}$ meters. Find its height after 6 seconds.
A) 90.0 meters
B) 125.2 meters
C) 102.96 meters
D) 97.2 meters

Question

Quizplus · Accepted Answer

Substitute $t = 6$ into the equation $s(t) = 21t - 0.8t^2$ to find the height: $s(6) = 21(6) - 0.8(6)^2 = 126 - 28.8 = 97.2$ meters.

Solve the Problem 21 m/s21 \mathrm{~m} / \mathrm{s}21 m/s , Its Height After ttt

Solve the Problem $21 \mathrm{~m} / \mathrm{s}$ , Its Height After $t$