homework6: problem 8\n(4 points)\na ball is shot straight up into the air from the ground with initial…
homework6: problem 8\n(4 points)\na ball is shot straight up into the air from the ground with initial velocity of 46 ft/sec. assuming that the air resistance can be ignored, how high does it go?\nhint: the acceleration due to gravity is -32 ft per second squared.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.\nemail instructor\npage generated october 19, 2025, 8:28:41 pm cdt\nwebwork © 1996 - 2024 | theme: math4 ttu | ww_version 2.19 | pg_version 2.19\nthe webwork project
Answer
Explanation:
Step1: Find the velocity function
The acceleration (a(t)=- 32). Integrating (a(t)) to get the velocity function (v(t)). Since (v(t)=\int a(t)dt), then (v(t)=\int - 32dt=-32t + C). Given the initial velocity (v(0) = 46), substituting (t = 0) into (v(t)): (v(0)=-32\times0 + C=46), so (C = 46). Thus, (v(t)=-32t + 46).
Step2: Find the time when the ball reaches the maximum height
At the maximum - height, the velocity (v(t)=0). Set (v(t)=-32t + 46 = 0). Solving for (t): (-32t=-46), then (t=\frac{46}{32}=\frac{23}{16}) seconds.
Step3: Find the position function
The position function (s(t)=\int v(t)dt). Since (v(t)=-32t + 46), then (s(t)=\int(-32t + 46)dt=-16t^{2}+46t + D). Given (s(0) = 0) (starts from the ground), substituting (t = 0) into (s(t)): (s(0)=-16\times0^{2}+46\times0+D = 0), so (D = 0). Thus, (s(t)=-16t^{2}+46t).
Step4: Find the maximum height
Substitute (t=\frac{23}{16}) into (s(t)): [ \begin{align*} s\left(\frac{23}{16}\right)&=-16\times\left(\frac{23}{16}\right)^{2}+46\times\frac{23}{16}\ &=-16\times\frac{529}{256}+\frac{1058}{16}\ &=-\frac{529}{16}+\frac{1058}{16}\ &=\frac{-529 + 1058}{16}\ &=\frac{529}{16}\ &=33.0625 \end{align*} ]
Answer:
(33.0625) feet.