Question

A projectile is shot upward from the surface of Earth with an initial velocity of 93 meters per second. Use the position function below for free-falling objects. What is its velocity after 4 seconds? (Round your answers to one decimal place.)s(t) = -4.92 + v0t + s0​

Answer 1

The velocity of the projectile after 4 seconds is 55.2 m/s.

Step-by-step explanation:

To find the velocity of the projectile after 4 seconds, we can use the position function for free-falling objects: s(t) = -4.9t^2 + v0t + s0.

Given that the initial velocity (v0) is 93 m/s, we can substitute this value into the position function to get s(t) = -4.9t^2 + 93t + s0. Since s0 is not given, we will assume it to be zero.

Therefore, the position function becomes s(t) = -4.9t^2 + 93t.

To find the velocity after 4 seconds, we will differentiate the position function with respect to time. Taking the derivative of s(t) gives us v(t) = -9.8t + 93.

Plugging in t = 4 into this equation, we can find the velocity after 4 seconds as v(4) = -9.8(4) + 93 = 55.2 m/s.