
#1
Sep1304, 05:44 PM

P: 33

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? 



#2
Sep1304, 05:49 PM

HW Helper
P: 2,280

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
Sep1304, 06:34 PM

Mentor
P: 40,890

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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