Answer:
at t = 1 s: +56 ft/s
at t = 3 s: -8 ft/s
Step-by-step explanation:
The equation that describes the heigth of the ball at time t is
[tex]s(t)=-16t^2+88t+4[/tex]
where
4 ft is the initial height at t = 0
+88 ft/s is the initial velocity of the ball
[tex]-32 ft/s^2[/tex] is the acceleration due to gravity
The instantaneous velocity of the ball can calculated as the derivative of the position.
Calculating the derivative of s(t) with respect to time, we find an expression for the instantaneous velocity:
[tex]v'(t)=\frac{ds(t)}{dt}=-2\cdot 16 t^{2-1} +88 t^{1-1}+4\\\rightarrow v(t) = -32t+88[/tex]
Now we can find the value of the instantaneous velocity at various times t:
- At t = 1 second, we have
[tex]v(1)=-32\cdot 1 + 88 = 56 ft/s[/tex]
- At t = 3 seconds, we have
[tex]v(3)=-32\cdot 3 + 88 =-8 ft/s[/tex]
