Calculating Freefall Time: Solving for t in a Ball's Descent

AI Thread Summary
To calculate the time it takes for a ball thrown downward at 8.25 m/s from a height of 29.4 m to hit the ground, the kinematic equation y = y_0 + v_0 t + (1/2)a t^2 should be used. It's crucial to maintain a consistent sign convention, treating both the acceleration due to gravity and the initial velocity as negative since they are directed downward. The final position when the ball strikes the ground is 0, leading to the equation 0 = 29.4 - 8.25t - 0.5(9.8)t^2. The variable for time should be squared in the equation, and solving for t will yield the correct time interval. Properly applying these principles will lead to the correct solution for the problem.
motionman04
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I'm having some problems with this question, A ball is thrown directly downward, with an initial speed of 8.25 m/s, from a height of 29.4 m. After what time interval does the ball strike the ground?

I tried 29.4 + 8.25 m/s(x) + 1/2(-9.8m/s)(x), but that didn't turn out to be right. Can I get some help with this one?
 
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Well in a xy coordinate system, initial position will be its height, and when it hits the ground it will have a position of 0, so it's final position must be 0
 
wrong sign... and more

motionman04 said:
I tried 29.4 + 8.25 m/s(x) + 1/2(-9.8m/s)(x), but that didn't turn out to be right. Can I get some help with this one?
I assume you are trying to apply the following kinematic equation:
y = y_0 + v_0 t + (1/2)a t^2
Be sure to use a consistent sign convention: not only is the acceleration negative (a = - 9.8 m/s^2), don't forget that the initial velocity is also negative since it is thrown downward.
And, as Cyclovenom points out, the final postion is where y = 0. Solve for t.
 
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29.4 + 8.25 m/s(x) + 1/2(-9.8m/s)(x),

Don't forget that the x (I would prefer t!) in the bold quantity needs to be squared.
 
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