homework6: problem 7\n(4 points)\na stone is thrown straight up from the edge of a roof, 825 feet above the…
homework6: problem 7\n(4 points)\na stone is thrown straight up from the edge of a roof, 825 feet above the ground, at a speed of 20 feet per second.\na. remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 6 seconds later?\nb. at what time does the stone hit the ground?\nc. what is the velocity of the stone when it hits the ground?\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\nemail instructor\npage generated october 19, 2025, 8:26:58 pm cdt\nwebwork © 1996 - 2024 | theme: math4 - ttu | ww_version: 2.19 | pg_version 2.19\nthe webwork project
Answer
Explanation:
Step1: Find the position function
The general formula for the position function (s(t)) of an object in vertical - motion is (s(t)=s_0 + v_0t+\frac{1}{2}at^{2}), where (s_0) is the initial position, (v_0) is the initial velocity, and (a) is the acceleration. Given (s_0 = 825) (initial height), (v_0=20) (initial velocity), and (a=-32) (acceleration due to gravity). So (s(t)=825 + 20t-16t^{2}).
Step2: Solve part A
For part A, we need to find (s(6)). Substitute (t = 6) into the position function: [ \begin{align*} s(6)&=825+20\times6-16\times6^{2}\ &=825 + 120-16\times36\ &=825+120 - 576\ &=369 \end{align*} ]
Step3: Solve part B
For part B, when the stone hits the ground, (s(t)=0). So we need to solve the quadratic equation (825 + 20t-16t^{2}=0). Using the quadratic formula (t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) for the quadratic equation (ax^{2}+bx + c = 0). Here (a=-16), (b = 20), and (c = 825). [ \begin{align*} t&=\frac{-20\pm\sqrt{20^{2}-4\times(-16)\times825}}{2\times(-16)}\ &=\frac{-20\pm\sqrt{400+52800}}{-32}\ &=\frac{-20\pm\sqrt{53200}}{-32}\ &=\frac{-20\pm20\sqrt{133}}{-32}\ &=\frac{20\pm20\sqrt{133}}{32}\ &=\frac{5\pm5\sqrt{133}}{8} \end{align*} ] We take the positive root (t=\frac{5 + 5\sqrt{133}}{8}\approx\frac{5+5\times11.53}{8}=\frac{5+57.65}{8}=\frac{62.65}{8}=7.83) (we discard the negative root since time (t\geq0)).
Step4: Solve part C
The velocity function (v(t)) is the derivative of the position function. Since (s(t)=825 + 20t-16t^{2}), then (v(t)=s^\prime(t)=20-32t). We use the time (t=\frac{5 + 5\sqrt{133}}{8}) from part B. [ \begin{align*} v\left(\frac{5 + 5\sqrt{133}}{8}\right)&=20-32\times\frac{5 + 5\sqrt{133}}{8}\ &=20-(20 + 20\sqrt{133})\ &=- 20\sqrt{133}\approx - 20\times11.53=-230.6 \end{align*} ]
Answer:
A. (369) feet B. (\frac{5 + 5\sqrt{133}}{8}\approx7.83) seconds C. (-20\sqrt{133}\approx - 230.6) feet per second