[SOLVED] Kinematics in two dimensions problem 1. The problem statement, all variables and given/known data Chapter 3, Problem 20 Incorrect. A car drives horizontally off the edge of a cliff that is 48.1 m high. The police at the scene of the accident note that the point of impact is 195 m from the base of the cliff. How fast was the car traveling when it drove off the cliff? 2. Relevant equations y=V(0)t+ (1/2)gt^2 y=(1/2)(V(0)+V)T- maybe? 3. The attempt at a solution Ok I think I've figured out the first part of the equation, finding out the time it took for the car to fly off the cliff- here's how I did it. y= V(0)t+ (1/2)gt^2 -48.1= (0)t +(1/2)(-9.8)t^2 t=3.13 Here's where I think I'm messing up somewhere: x=195 (how far the car flew off the cliff) t=3.13 (what i got from the previous answer) V(0)= ? (what i need to solve for) V= 0 ( the car doesn't have any more veloicty after it crashed into something) however, when i plug this into my equation : x=(1/2)(V(0)+V)t, it says i didn't get the right answer. here's my math just for a double check 195=(1/2)(V(0)+0)(3.13) 195=1.57V(0) V(0)= 124.5 Any thoughts?