The distance is the hypotenuse of a right triangle where one leg is the x distance and the other leg is the y distance. In this case, the y values are the same so the distance between the points is simply the x distance.
A more complete explanation... The car is traveling horizontally at a constant speed. When it drives off the cliff, it keeps going at the same speed (disregarding friction) until it is stopped by hitting the ground. A forumla for (d) distance traveled with constant acceleratio...
#1 correct #2 correct #3 wrong #4 correct #5 wrong ---- #6 find t in seconds: 122.5m = (0.5)(g)(t^2) 122.5m = (0.5)(9.8m/s^2)t^2 reducing... 25s^2 = t^2 t = +5 seconds and t = -5 seconds Disregard the negative time. That is how long it took to fall 122.5m. So, if the car trave...
Be careful. You posted (72)/(-68) reduces to (-18)/(-17). You had a sign change in the numerator. Probably just a typo, but you might want to check.
The slope is 2.
In the equation s(t) = (s0) + (v0)t + (1/2)(a)(t^2) s(t) gives the height for some time t. s0 is the initial height, which is the height of the cliff. v0 is the initial velocity, for this problem 20m/s. The last term is the distance due to gravity. As t^2 increases, this term ...
You have the right formula. Let S0 be the height of the cliff in meters. When the stone stops it is at 0m height. So, 0m = S0 + (v0)t + (1/2)gt^2 t =7 s g=-9.8m/(s^2) 0 = S0 + (20m/s)(7s) + (1/2)(-9.8m/(s^2))(49s^2) 0 = S0 + 140m - 240.1m S0 = 100.1m
You have an error in your first step when you multiplied: -2(5 + 4a) It is not -10 + 8a .
Do you see a numeric factor that is common in each term? Factor that number from each term. What is left?
v1 = velocity one second later vi = initial velocity of rock = 4.9 m/s t = time in seconds g = acceleration due to gravity = -9.8 m/s^2 v1 = vi + gt Plug the numbers in and solve for v1.