MHB How Long Until the Ball Hits the Ground?

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To determine when the ball hits the ground, the problem involves using the kinematic equation x = x0 + v0t + at²/2, with given values of initial height (1m), initial velocity (+10 m/s), and acceleration due to gravity (−9.81 m/s²). The equation simplifies to 0 = 1 + 10t - 4.905t², which is a quadratic equation. The discussion highlights the need to apply the quadratic formula to solve for time, rather than isolating the variable t directly. Participants emphasize the importance of maintaining proper algebraic operations when manipulating the equation. The solution will yield the time it takes for the ball to reach the ground.
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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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