Physics Question.. please show work..thank you?

Question

A parachutist descending at a speed of 15.0 m/s loses a shoe at an altitude of 30.0 m. (Assume the positive direction is upward.)
(a) When does the shoe reach the ground?
s
(b) What is the velocity of the shoe just before it hits the ground?
m/s

Answer 1

Assuming that the shoe's Initial velocity equals -15 m/s and h=30.0m :
(a)
S=(Vi)(t) + 1/2 (a)(t)^2 best formula for now.
-30=(-15)(t) + 1/2(-9.8)(t)^2 {s=(-30) because its falling down}
-30= (-15t) + ( -4.9)(t)^2 {Vi= (-15) because its falling down}
0= -4.9(t)^2 - 15t + 30 {Vi= Initial Velocity}

t= 15 / [2(-4.9)] ± [√(-15)^2 - 4(-4.9)(30)] / [2(-4.9)]
t= -1.5306 ± [√225 + 588] / (-9.8)
t= -1.5306 ± [28.5132] / (-9.8)
t= -1.5306 ± [-2.9095]
t= 1.3789 {ignore the negative result, cause time ≠ negative}

(b)
Vf= Vi + gt
Vf= -30 + (-9.8)(1.3789)
Vf= -43.5132 m/s

Hope this helps =)

Answer 2

Postion = y = 1/2at^2 + v0t +y0

Now a = -g = 10 m/s^2, v0 = -15 m/s, and y0 =30 m and y = 0

0 = -5t^2 - 15 t +30

0 = t^2 + 3t - 6 ---> t = -3/2 +/-sqrt(9 +24) /2 = -3/2 +/-sqrt(33)/2

t = 4. 37, 1.37 sec --> 4.37 is too big so t =1.37 s

v = -gt-v0t = -34.3 m/s