A rock that is thrown upward from a bridge with an initial velocity of 2.0 m/s. (Assume there is no air drag: a = -10m/s2).
a) If the rock takes 4.0 s to hit the water below once the rock is released. What is the final velocity of the rock as it hits the water?
b) How high is the bridge?
v = u +at
x = x0 + vit + 1/2*a*t2
The Attempt at a Solution
For (a), I know that the equation I can use is v = u + at. u being initial velocity.
v = (2.0m/s) + (-10m/s2) (4.0s)
v = -38 m/s
However, for (b), I do not know how I can solve it and need help on that ASAP