byerskm2 Group Title An object is dropped from a tower of height 224 feet, subject to only the constant acceleration of −32 ft/sec2 due to gravity. (d) What should its initial velocity be in order to strike the ground after 1 second? one year ago one year ago

1. satellite73 Group Title

clearly the word "dropped" here is inappropriate, since they are asking what hthe initial velocity should be if it is dropped, the initial velocity is 0!!

2. byerskm2 Group Title
3. byerskm2 Group Title

In order to reach the ground in 1 second the Vi would have to have a negative velocity of some value otherwise it would not be fast enough

4. satellite73 Group Title

oh you put in the word "dropped" tsk tsk

5. byerskm2 Group Title

I just can't find the corriect value

6. byerskm2 Group Title

sorry that's what the question said

7. byerskm2 Group Title

:P

8. satellite73 Group Title

lets try this $a(t)=-32, v(t)=-32t^2+v_0, p(t)=-16t^2+v_0t+224$

9. byerskm2 Group Title

yes

10. satellite73 Group Title

or more commonly written $h(t)=-16t^2+v_0t+224$

11. satellite73 Group Title

set equal to zero, then put $$t=1$$ to solve for $$v_0$$

12. byerskm2 Group Title

208?

13. satellite73 Group Title

$t^2-\frac{v_0}{16}-14=0$

14. satellite73 Group Title

oh no your answer will have a $$v_0$$ in it

15. byerskm2 Group Title

I plugged in 1 for t and got the 208

16. satellite73 Group Title

oh wait, you are right put $$t=1$$ set equal to 0 solve for $$v_0$$

17. satellite73 Group Title

$0=-16+v_0+224$ $v_0=-208$

18. satellite73 Group Title

i see the problem it is negative and you put 208 i bet

19. byerskm2 Group Title

Oh! Darn. I even said it have to be negative in the beginning didn't I? X(

20. satellite73 Group Title

yes you did

21. byerskm2 Group Title

I plugged it in :) it's right :P thanks! I knew I had to be right but I'm always missing some stupid detail!

22. satellite73 Group Title

yw always good to have another pair of eyes