a ball is thrown vertically upward from the top of a 100 foot tower, with an initial velocity of 20ft/sec. Its position function is s(t)=-16t^2+20t+100. What is its velocity in ft/sec when t=1 second

Respuesta :

we are given

position function as

[tex]s(t)=-16t^2+20t+100[/tex]

Since, we have to find velocity

so, we will find derivative with respect to t

[tex]s'(t)=-16*2t+20*1+0[/tex]

[tex]s'(t)=-32t+20[/tex]

[tex]v(t)=-32t+20[/tex]

now, we can plug t=1

[tex]v(1)=-32*1+20[/tex]

[tex]v(1)=-12ft/sec[/tex].................Answer