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(52798) About Me  (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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    - 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

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Answer by math_tutor2020(3817) About Me  (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
t+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

t+=+%28-%28-20%29%2B-sqrt%28%28-20%29%5E2-4%284.9%29%2820%29%29%29%2F%282%284.9%29%29

t+=+%2820%2B-sqrt%288%29%29%2F%289.8%29

t+=+%2820%2Bsqrt%288%29%29%2F%289.8%29 or t+=+%2820-sqrt%288%29%29%2F%289.8%29

t+=+2.329431 or t+=+1.752201
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.