The Rock's Height and Velocity: 16t-1.86t^2

[tc903]

H=16t-1.86t^2 is the height formula. The ball was thrown 16 m/s.

I derived 16-3.72a.

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?
View attachment 4471This is what I chose.

[-4.-2)
(-2,2)
[2,4)
(4,6)
(6,8)

 

[Sudharaka]

H=16t-1.86t^2 is the height formula. The ball was thrown 16 m/s.

I derived 16-3.72a.

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 $H=16t-1.86t^2$ we have, $\frac{dH}{dt}=16-3.72t$. The rock reaches the ground at $t=8.6$ and therefore the velocity is, $\left.\frac{dH}{dt}\right|_{t=8.6}=16-3.72(8.6)=-15.992$.

 

[tc903]

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.