Recent content by mackenzieb

I don't understand what you mean by that so I do need to incorporate the angle into this question ...meaning the initial velocity is still 0 but is 35 degrees in the y-direction? I don't know how I am supposed to place this into the equation

Yes that is right, the object was thrown with an initial upward angle of 35 degrees. I am trying to use the examples in my notebook I have been provided with, and that it what is making this so difficult. I didnt think the angle had to be applied until the second question, but i must be...

here is my complete work so far: Given: d=4.0m [down] a=9.8m/s^2 [down] vi=0m/s (whenever something falls from a height, the v1 is always 0) Required: t vf Therefore, I use the equation d=vi(t)+1/2a(t)^2 4m [down] = 0m/s(t)+1/2 (9.8m/s^2[down])(t^2)...

when i get to the final step... -2.86+/-sqrt-2.86^2-4(4.906)(-4)/2(4.906) i get -2.86+/- 2.677007231 i am unsure what to do with the -2.86+/-

so when i use that equation...i get t=-2.86+/- sqrt -2.86^2-4(4.906)(-4)/2(4.906)

okay so to make sure I am onthe right track then... a=4.906 b=2.86 c=-4 -4m=(2.86m/s)t+1/2(-9.8m/s^2)t^2 1/2(9.8m/s^2)t^2-(2.86m/s)t-4m=0 -2.86 +/- sqrt 2.86^2-2(4.906)(4)/2(-4/906) is that the correct setup?

does that make the a term -4.9? (9.8 divided by 1/2)

there is no negative sign, therefore 9.8 should remain positive. making the variables: a=9.8 b=2.86 c=-4

if i am correct.. a=-9.8 (not sure if this is supposed to be -, but i believe so) b=2.86 c=-4

I used this reply as reference "hey Phee, go back to my post if you want. I explained that you should use the quadratic formula. as long as 0=ax^2 + bx + c then you can solve for x using a particular quadratic formula so if 0 = 1/2(-9.81m/s^2)t^2 + 2.86 m/s t + 4m you can solve for t using...

-------------------------------------------------------------------------------- I am still getting 1.07 s as my answers even after putting the acceleration due to gravity as a negative. I keep getting answers similar to 1.07 s which i know is not correct. This is how i have been doing the...

I also am having troubles with this problem. I keep getting answers similar to 1.07 s which i know is not correct. This is how i have been doing the problem: dv=viyXt+1/2(9.8m/s)^2(t)^2 -4m=(2.86m/s)(t)+1/2(9.8m/s)^2(t)^2 t=square root of: t (-4) / -9.8+2.86 =1.07 s Can someone please...