Question

A projectile is shot upward from the surface of Earth with an initial velocity of 106 meters per second. Use the position function below for free-falling objects. What is its velocity after 5 seconds? After 13 seconds? (Round your answers to one decimal place.) s(t) = −4.9t2 + v0t + s0

Answer 1

Answer:

• 5 seconds: 57 m/s
• 13 seconds: -21.4 m/s

Step-by-step explanation:

The velocity of the object is the derivative of its position, so is ...

  s'(t) = -9.8t +v0 = 106 -9.8t

At the given times, the object's velocity is ...

  s'(5) = 106 -9.8·5

  s'(5) = 57 . . . . m/s

  s'(13) = 106 -9.8·13

  s'(13) = -21.4 . . . . m/s

After 5 seconds, its velocity is 57 m/s; after 13 seconds, -21.4 m/s.