Question

An object is released from rest and falls a distance H during the first second of time.  How far will it fall in the third second of time?

		H
		4H
		5H
		9H
		H2

Answer 1

So answer is 5H

Answer Given, initial speed, u=0 distance travell in first second =H distance travell in third second = H, (say) We have the equation of motion, distance travelled in nth secona, Sn = utta (2n-1) where a is acceleration. Hence we can write, H H ut ťa (2x 8-1) utta (241.-1) 'Hi H of Ixax 5 of Ixa X1 => HI 비 Hi = 5 H Ans. so in the third second of time the object will fall by the distance 5H