SOLUTION: An object is propelled vertically upward with an initial velocity of 20 meters per second. The distance s (in meters) of the object from the ground after t seconds is given by s =
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Question 1208731: An object is propelled vertically upward with an initial velocity of 20 meters per second. The distance s (in meters) of the object from the ground after t seconds is given by s = -4.9t^2 + 20t.
A. When will the object be 15 meters above the ground?
For part A, I must let s = 15 and solve for t. You say?
B. When will the object strike the ground?
For part. B, I must let s = 0 and solve for t. Correct?
C. When will the object reach a height of 100 meters?
For part C, let s = 100 and solve for t.
Am I correct?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52799) (Show Source): You can put this solution on YOUR website!
.
Yes, you must solve it.
Especially for part (C), because when you try to do it,
you will discover other issues, that will require your analysis.
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If you want to learn the subject to that level to solve such problems on your own
and to learn it from a good source, you have this lucky, happy and fortunate opportunity.
In this site, I prepared the lessons that cover the entire subject
- Introductory lesson on a projectile thrown-shot-launched vertically up
- Problem on a projectile moving vertically up and down
- Problem on an arrow shot vertically upward
- Problem on a ball thrown vertically up from the top of a tower
- Problem on a toy rocket launched vertically up from a tall platform
- A flare is launched from a life raft vertically up
Consider these lessons as your textbook, handbook, tutorials and (free of charge) home teacher.
Read them attentively and learn how to solve this type of problems, how to analyze and how to present a solution, once and for all.
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
All three statements are correct.
After the substitution, you can use the quadratic formula to find t.
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Example question: when will the object be 20 meters above the ground?
Solution:
s = -4.9t^2 + 20t
20 = -4.9t^2 + 20t
4.9t^2 - 20t + 20 = 0
Plug a = 4.9, b = -20, c = 20 into the quadratic formula
or
or
Both decimal results are approximate.
The object reaches the 20 meter mark at around 1.752 seconds, rises higher a little bit, then falls back down to the 20 meter mark at around 2.329 seconds
Another approach is to graph the parabola y = -4.9x^2 + 20x and the horizontal line y = 20
The two intersect at points A and B shown in the graph above.
The x coordinates of these intersections are the approximate solutions found earlier.
GeoGebra and Desmos are two graphing options, among many others.
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