How Long Until the Ball Hits the Ground?

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In summary, the problem is asking for the time it takes for a ball thrown straight up with an initial velocity of +10 m/s and starting at a height of 1m to hit the ground, given the acceleration due to gravity of -9.81m/s^2. The appropriate kinematic equation to use is x = x0 + v0t + at^2/2. After plugging in the given values and simplifying, we get the equation 0=1+10t-4.905t^2, which can be solved using the quadratic formula.
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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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  • #2
Where exactly are you getting stuck? Your equation is correct, but can you show what you have tried so far?
 
  • #3
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
 
  • #4
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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  • #5


Based on the given information, the kinematic equation you should use is x = x0 + v0t + 1/2at^2, where x0 is the initial height (1m), v0 is the initial velocity (+10 m/s), and a is the acceleration due to gravity (-9.81m/s^2). The target variable, as you correctly identified, is time (t).

To solve for the time when the ball hits the ground, we can set x = 0 (since the ground is at a height of 0) and solve for t. This gives us the following equation:

0 = 1 + 10t - 4.905t^2

Solving for t using the quadratic formula, we get two solutions: t = 0.204s and t = 2.047s. However, since the initial velocity is positive (+10 m/s), we can discard the negative solution and conclude that the ball will hit the ground after 2.047 seconds.

Therefore, the answer to the problem is that the ball will hit the ground after 2.047 seconds.
 

FAQ: How Long Until the Ball Hits the Ground?

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