Freefall physics homework

  • #1
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?
 

Answers and Replies

  • #2
Pyrrhus
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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
 
  • #3
Doc Al
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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.
 
Last edited:
  • #4
Integral
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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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