The problem asks for the velocity of a projectile at two different times (2 seconds and 11 seconds) when it is launched upward with an initial velocity of 119 m/s. The provided position equation is:

$s(t) = -4.9t^2 + v_0t + s_0$

where:

• $v_0 = 119 \, \text{m/s}$ is the initial velocity,
• $s_0 = 0 \, \text{m}$ (since the projectile is shot from the surface of the Earth),
• $-4.9t^2$ accounts for the gravitational acceleration ($g = 9.8 \, \text{m/s}^2$).

Step 1: Deriving the velocity equation

Velocity is the derivative of the position with respect to time:

$v(t) = \frac{d}{dt} \left( -4.9t^2 + v_0t + s_0 \right) = -9.8t + v_0$

Since $v_0 = 119 \, \text{m/s}$, the velocity function becomes:

$v(t) = -9.8t + 119$

Step 2: Calculating the velocity at 2 seconds

For $t = 2$ seconds:

$v(2) = -9.8(2) + 119 = -19.6 + 119 = 99.4 \, \text{m/s}$

Step 3: Calculating the velocity at 11 seconds

For $t = 11$ seconds:

$v(11) = -9.8(11) + 119 = -107.8 + 119 = 11.2 \, \text{m/s}$

Final answers:

• The velocity at 2 seconds is 99.4 m/s.
• The velocity at 11 seconds is 11.2 m/s.

Would you like more details on the steps or concepts used?

Here are some related questions you might consider:

1. How does the negative sign in the velocity equation relate to the projectile's motion?
2. What happens to the velocity after the projectile reaches its peak height?
3. How would the initial velocity impact the maximum height of the projectile?
4. What is the significance of the velocity being close to zero at 11 seconds?
5. How can we find the time at which the projectile reaches its maximum height?

Tip: To find the maximum height of a projectile, set its velocity equal to zero and solve for the time at which this occurs.