please help me with calc

Question

A projectile is shot upward from the surface of Earth with an initial velocity of 111 meters per second. Use the position function below for free-falling objects. What is its velocity (in m/s) after 3 seconds? After 14 seconds? (Round your answers to one decimal place.)

s(t) = −4.9t² + v₀t + s₀

3 s m/s:

14 s m/s:

Answer 1

s(t) = -4.9t^2 + 111vo + so

v(t)= s'(t) = -9.8t + 111

v(3) = 111-29.4 = 81.6 meters per second after 3 seconds

v(11) = 111-9.8(11) = 3.2 meters per second after 11 seconds

at 11 seconds it's amost reached maximum height

where velocity becomes zero for an instant in time

s(3) =-4.9(3)^2 +111(3)

= -44.1 +333 = 288.9 meters after 3 seconds

h(11) =-4.9(11)^2 + 111(11)

= -592.9 + 1221 = 628.1 meters after 11 seconds

Answer 2

v = ds/dt = -9.8t + 111

Plug in the t values for v at any time.