Question

bob drop a rock from a height of 4.5 m. (the acceleration due to gravity is 9.8 m/s^2) 1.at s the initial speed of the rock 2. how long is the rock falling before it hits the ground c. what is the final speed of the rock? [I have trouble with c)

My work:
a) v1 =0m/s
b) v1t +1/2at^2
0t + 1/2 (-9.8) t^2
-4.5=1/2(-9.8)t^2
square of root of -4.5/-4.9 =t
0.95s=t
c) v2=?
v2=v1 + at
=0 t (-9.8)(0.96)
=-9.408 m/s [I got this answer wrong, my answer sheet says I should get positive 9.408]

Answer 1

a. Correct

b. h = g*t^2/2 = 4.5
4.9*t^2 = 4.5
t^2 = 0.9184
t = 0.958 s.

c. V = Vo + g*t = 0 + 9.8*0.958 = 9.39 m/s.

or V^2 = Vo^2 + 2g*d = 0 + 19.6*4.5 = 88.2
V = 9.39 m/s.

or V^2 = 0 + (-19.6*(0-4.5) = 88.2
V = 9.39 m/.s.