when a stone is dropped from a building 200 m high , its speed is proportional to the time elapsed after dropping . the distance traveled is proportional to the square of the time elapsed . after 1 second the speed of the train was 10 m / sec and it was 190 m above the ground . when its speed is 25 m / sec , what would be its distance from the ground ?
v = x * t and s = y * t ^ 2 10 = x * 1 , hence x = 10 units ( 200 - 190 ) = y * 1 * 1 y = 10 when speed = 25 m / sec , time elapsed = 25 / 10 = 2.5 sec distance travelled by stone = 10 * ( 2.5 ) ^ 2 = 62.5 mts so distance of stone above ground = 200 - 62.5 = 137.5 mtrs answer : b
subtract(200, multiply(power(divide(25, subtract(200, 190)), const_2), 10))