A car drives straight off the edge of a cliff that is 55 m high. The police at the scene of the accident observed that the point of impact is 121 m from the base of the cliff. How fast was the car traveling when it went over the cliff?
2006-10-01 16:41:25 · 3 answers · asked by Alan l 1 in Education & Reference ➔ Homework Help
You need to find out how long it took the car to fall 55 m.
x(t) = 0.5*g*t^2 + v0 * t + x0.
g = -9.8 m/s^2
v0 = 0
x0 = 55.
Find t.
Once you have t then the car's velocity was 121m/t.
2006-10-01 16:55:07 · answer #1 · answered by feanor 7 · 0⤊ 0⤋
The car was traveling 36.13 m/s
I think this is right... using the equation x = (V initial)(time) + (1/2)(a)(t)^2 we can find the total time the car was falling. If we think that we just drop the car from no initial V, then the equation becomes x = (1/2)(a)(t)^2 Solve this for T: X=55m A=gravity (9.8) therefore T= 3.35 sec.
In physics the x component is independent and dose not depend on the y.... so in English, in a perfect world (with no air resistance) the car will continue traveling at the speed it was going when it left the cliff until the time it hits the ground.
Using the time the car was falling (3.35 sec) we can calculate the velocity of the car using the equation V = x / T (x = distance T = time) From that, the velocity of the car was 36.13 m/s when it left the cliff to land 121m from the base of the cliff.
2006-10-02 00:10:08 · answer #2 · answered by cardss1 1 · 0⤊ 0⤋
falling objects (starting at zero speed) fall at a rate of 10 meters per second, squared.
2006-10-02 00:12:01 · answer #3 · answered by doe 3 · 0⤊ 0⤋