The given problem describes a vertically projected object, starting from a height of 4 meters, with an initial velocity of $25.5 \, \text{m/s}$, where the height $h$ in meters after $t$ seconds is represented by the equation:

$h(t) = 4 + 25.5t - 4.9t^2$

Here's a breakdown of each part of the problem:

1. Velocity after 2 and 4 seconds
   You correctly calculated the velocities as:

   • $v(2) = 5.9 \, \text{m/s}$
   • $v(4) = -13.7 \, \text{m/s}$

2. Maximum Height The time at which the projectile reaches the maximum height is $t = 2.602 \, \text{s}$ (correct). At this time, the maximum height is calculated to be $37.18 \, \text{m}$.

3. Time when it hits the ground The projectile hits the ground after $t = 5.36 \, \text{s}$ (correct).

4. Impact Velocity
   The velocity when it hits the ground is incorrectly calculated. Here's the correct approach:

   To find the velocity, we use the velocity equation derived from the height equation. The velocity $v(t)$ is the derivative of the height equation $h(t)$:

   $v(t) = \frac{d}{dt} \left( 4 + 25.5t - 4.9t^2 \right) = 25.5 - 9.8t$

   Now, substitute $t = 5.36 \, \text{s}$:

   $v(5.36) = 25.5 - 9.8(5.36) = 25.5 - 52.528 = -27.028 \, \text{m/s}$

   So, the velocity when it hits the ground is approximately $-27.03 \, \text{m/s}$, not $-26.254 \, \text{m/s}$.

Would you like more details on any part of this? Here are some follow-up questions for deeper understanding:

1. How is the derivative used to find velocity from a position function?
2. Why does the velocity become negative as time progresses?
3. How do we find the time when the projectile hits the ground using the quadratic formula?
4. How does air resistance affect the real-life scenario of a projectile's motion?
5. Can we calculate the acceleration at any point in time during the projectile's motion?

Tip: When working with projectile motion, always check the signs of velocity, especially after the object reaches its maximum height, as it switches from positive (upward motion) to negative (downward motion).