
#1
June 23rd, 2015,
12:23
H=16t1.86t^2 is the height formula. The ball was thrown 16 m/s.
I derived 163.72a.
When will the rock hit the surface. It hits the surface 8.6 seconds.
What velocity does the rock hit the surface? H=0.0344 then the velocity is 15.87 m/s?
This is what I chose.
[4.2)
(2,2)
[2,4)
(4,6)
(6,8)
Last edited by tc903; June 23rd, 2015 at 13:28.

June 23rd, 2015 12:23
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#2
June 25th, 2015,
23:02
Originally Posted by
tc903
H=16t1.86t^2 is the height formula. The ball was thrown 16 m/s.
I derived 163.72a.
When will the rock hit the surface. It hits the surface 8.6 seconds.
What velocity does the rock hit the surface? H=0.0344 then the velocity is 15.87 m/s?
This is what I chose.
[4.2)
(2,2)
[2,4)
(4,6)
(6,8)
Hi tc903,
I think the velocity is wrong. Since $H=16t1.86t^2$ we have, $\frac{dH}{dt}=163.72t$. The rock reaches the ground at $t=8.6$ and therefore the velocity is, $\left.\frac{dH}{dt}\right_{t=8.6}=163.72(8.6)=15.992$.

#3
June 26th, 2015,
00:19
Thread Author
I would think the same, maybe I shouldn't have rounded it to 15.9 when I originally did that way. I have a computer that is telling me no when I plug in an answer like that.