   SEARCH HOME Math Central Quandaries & Queries  Question from Mimi, a student: A car fails to take a sharp turn on a flat road and plunges off a cliff. The car lands 65m horizontally and 43m vertically from where it left the edge of the cliff. How fast was the car travelling? Hi Mimi.

The car fell 43m. It was a flat road, so the vertical speed of the car was initially zero. Therefore the time the car took to fall is entirely due to gravity.

To find out how fast the car was going, you need to realize that it was going that speed horizontally from the time it left the road until it crashed 65m horizontally away. If you knew how long it took to fall, you could calculate the speed it must have been going to move 65 m in that time.

To find the time, first remember that the acceleration due to gravity is 9.8 m/s2. Thus, the distance fallen is d = (1/2) gt2, where d is the distance fallen (43 m), g is the acceleration due to gravity (9.8 m/s2) and t is the time. So you have
43 = (1/2) 9.8 t2

Solve this for t, then calculate its value. Use that to find the speed necessary to go 65 meters in that time.

Cheers,
Stephen La Rocque     Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.