MHB How Long Until the Ball Hits the Ground?

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The discussion focuses on calculating the time it takes for a ball, thrown upwards with an initial velocity of +10 m/s from a height of 1 meter, to hit the ground under the influence of gravity at -9.81 m/s². The correct kinematic equation to use is x = x0 + v0t + at²/2, which simplifies to a quadratic equation: 0 = 1 + 10t - 4.905t². Participants emphasize the application of the quadratic formula to solve for time, highlighting the importance of maintaining consistent units throughout the calculation.

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Problem Statement: A boy throws a ball straight up into the air with an initial velocity of +10 m/s, from a height of 1m. Note that the acceleration due to gravity is −9.81m/s2. When does the ball hit the ground?

So I know that the Target Variable is Time, however I'm not sure which of the four kinematic equations I should use.

I thought I would use; x = x0 + v0t + at2/2, but I keep getting stuck.

Any help would be appreciated, thanks.
 
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Where exactly are you getting stuck? Your equation is correct, but can you show what you have tried so far?
 
Sure, after plugging in all the givens I got this; 0=1m+10m/s(time)+[(-9.81m/s^2)(time^2)/2]

Now when I try to isolate the target variable, time, I get incoherent stuff like this; 2(-1m/(10m/s)*(-9.81m/s^2))=t+t^2
 
If you divide by 10 on one side, you need to divide by 10 on the other side too, so that's incorrect. You don't need to isolate the variable t, though. You already have it in a form where the quadratic formula can be applied to it:
$$0=1+10t- \frac{9.81}{2}t^2 = 1+10t-4.905t^2.$$
(The units cancel out, so all the terms have dimensions of length.) Have you been introduced to the quadratic formula yet?
 
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