Question

Vertical Motion In Exercises 99 and \(100,\) use the position function \(s(t)=-4.9 t^{2}+v_{0} t+s_{0}\) for free-falling objects. A projectile is shot upward from the surface of Earth with an initial velocity of 120 meters per second. What is its velocity after 5 seconds? After 10 seconds?

Short answer

The velocity of the projectile after 5 seconds is 71m/s, and after 10 seconds it is 22m/s.

Step-by-step solution

Derive the Position Function

The velocity equation would be the derivative of the position function. So, differentiate the given function \(s(t)=-4.9 t^{2}+v_{0} t+s_{0}\) with respect to time \(t\). This will give the velocity function \(v(t)\), which is \(v(t) = -9.8t + v_0\).

Calculate the Velocity at t=5 seconds

Substitute \(t=5\) and \(v_0=120\) into the velocity equation. This gives \(v(5) = -9.8(5) + 120 = -49 + 120 = 71m/s\). Thus, the velocity after 5 seconds is 71m/s.

Calculate the Velocity at t=10 seconds

Substitute \(t=10\) and \(v_0=120\) into the velocity equation. This gives \(v(10) = -9.8(10) + 120 = -98 + 120 = 22m/s\). Thus, the velocity after 10 seconds is 22m/s.