#### Answer

(a) After 2 seconds, the velocity of the projectile is $4.9~m/s$
After 4 seconds, the velocity of the projectile is $-14.7~m/s$
(b) The projectile reaches maximum height after $2.5$ seconds.
(c) The maximum height is $32.6$ meters
(d) The projectile hits the ground after $5.08$ seconds.
(e) When the projectile hits the ground, the velocity is $-25.3~m/s$

#### Work Step by Step

(a) $h = 2+24.5t-4.9t^2$
$v = 24.5-9.8t$
After 2 seconds:
$v = 24.5-9.8(2) = 4.9$
After 2 seconds, the velocity of the projectile is $4.9~m/s$
After 4 seconds:
$v = 24.5-9.8(4) = -14.7$
After 4 seconds, the velocity of the projectile is $-14.7~m/s$
(b) When the projectile reaches the maximum height, the velocity is 0.
$v = 24.5-9.8t = 0$
$t = \frac{24.5}{9.8}$
$t = 2.5$
The projectile reaches maximum height after $2.5$ seconds.
(c) $h = 2+24.5t-4.9t^2$
$h = 2+24.5(2.5)-4.9(2.5)^2$
$h = 32.6$
The maximum height is $32.6$ meters
(d) $h = 2+24.5t-4.9t^2 = 0$
$4.9t^2-24.5t-2 = 0$
We can use the quadratic formula to find the time $t$ when the projectile hits the ground:
$t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$t = \frac{-(-24.5) \pm \sqrt{(-24.5)^2-4(4.9)(-2)}}{(2)(4.9)}$
$t = \frac{24.5 \pm \sqrt{600.25+39.2}}{9.8}$
$t = -0.08~~$ or $~~t = 5.08$
The projectile hits the ground after $5.08$ seconds.
(e) $v = 24.5-9.8t$
$v = 24.5-9.8(5.08) = -25.3$
When the projectile hits the ground, the velocity is $-25.3~m/s$