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Author Question: Ron throws a ball straight up with an initial speed of 40 feet per second from a ... (Read 208 times)

vHAUNG6011

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on: Feb 28, 2021
Solve the problem.

Ron throws a ball straight up with an initial speed of 40 feet per second from a height of 7 feet. Find parametric equations that describe the motion of the ball as a function of time. How long is the ball in the air? When is the ball at its maximum height? What is the maximum height of the ball?
◦ x = 0, y = -16t2 + 40t + 7
2.311 sec, 1.25 sec,
6.988 feet
◦ x = 0, y = -16t2 + 40t + 7
4.621 sec, 1.25 sec,
177.892 feet
◦ x = 0, y = -16t2 + 40t + 7
5.328 sec, 1.25 sec,
25 feet
◦ x = 0, y = -16t2 + 40t + 7
2.664 sec, 1.25 sec,
32 feet
Precalculus ¦ Sullivan ¦ 10th Edition
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Marked as best answer by vHAUNG6011 on Feb 28, 2021

mariahkathleeen

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Reply #1 on: Feb 28, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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vHAUNG6011

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Reply 2 on: Feb 28, 2021
Thanks for the timely response, appreciate it

tandmlomax84

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Reply 3 on: Yesterday
Wow, this really help

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