Projectile Height Calculation: Time to Reach 10m with Given Initial Velocity

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The discussion revolves around solving a physics problem related to projectile motion, specifically calculating the time it takes for a projectile to reach a height of 10 meters with an initial vertical velocity of 20 m/s. The correct formula for height is provided, but the user initially makes an error in rearranging the equation, leading to incorrect coefficients in the quadratic equation. The correct formulation should be 9.8t^2 - 40t + 20 = 0, which allows for the calculation of the two valid times: 0.58 seconds and 3.50 seconds. The user acknowledges the mistake and expresses relief upon identifying the error. Understanding the correct application of the projectile motion formula is crucial for accurate calculations.
Bartosz
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Well its been a while since I've done any physics problems and I am having a problem with a real simple problem. The problem is:

You know that the height (in m) of a projectile is given by the formula

h = - g * t^2 / 2 + vi * t
where t is the time from launch, g is the acceleration due to gravity (9.8 m/sec2) and vi is the initial vertical velocity component (in m/sec).

Calculate the time at which a projectile will reach a certain height.

vi = 20
desired height = 10

So far i have this done, but i always get a negative number and the answers should be 0.58 sec and 3.50 sec.

10 = -9.8t^2/2 + 20t

10(2 + 20t) = -9.8t^2
20 + 200t = -9.8t^2

9.8t^2 + 200t + 20 = 0

(-200 + sqrt(200^2 - 4(9.8*20)) / 2(9.8)
-200 + sqrt(40000 - 784)) / 19.6
(-200 + 198.0303)/19.6

Im sure its just a stupid math mistake, but i can't find it. Also where we even given the correct equation?
 
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10 = -9.8t^2/2 + 20t
<={this is the step where your mistake is done}=>
10 - 20t = -9.8t^2/2
<==>
20 - 40t = -9.8t^2
<==>
9.82t^2 - 40t + 20 = 0
and not 9.8t^2 + 200t + 20 = 0
 
Thanks a lot, wow don't i feel stupid now :-p
 
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