A ball is thrown vertically upward from the top of a 200 foot tower, with an initial velocity of 5 ft/sec. Its position function is s(t) = –16t2 + 5t + 200. What is its velocity in ft/sec when t = 3 seconds?

Respuesta :

v = ds/dt = -32t + 5

at t = 3  velocity = -96 + 5  = -91 ft/sec answer

Answer:

The velocity would be - 91 ft/sec.

Step-by-step explanation:

Given,

The function that shows the position of the ball after t seconds,

[tex]s(t) = -16t^2 + 5t + 200[/tex]

Since, velocity is the changes in position with respect to time,

That is, if v(t) is the velocity of the ball after t second,

[tex]\implies v(t)=\frac{d}{dt}(s(t))[/tex]

[tex]=\frac{d}{dt}(-16t^2 + 5t + 200)[/tex]

[tex]=-32t+5[/tex]

Hence, the velocity after 3 seconds is,

[tex]v(3)=-32(3)+5=-96+5=-91\text{ ft per seconds}[/tex]