Question

A body is projected vertically upwards so that so that it travels same distance in 10th and 11th second of the journey the velocity of projection is (g=10m/s2)

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Answer 1

Answer:

Let u be the initial velocity of the ball. Distance travelled in the nth sec is given by,

Sn = u+1/2​×a×(2n–1)

Distance covered in 10th second:

(put a = -9.8 and n = 10)

S10 ​= u - 93.1

Distance covered in the 11th second:

(put n =11, a=-9.8 in equation):

S11 ​= u – 102.9

Now, as the body is projected upwards, the distance travelled in two different time intervals will be same only when at one time it is moving upwards and the other time it is coming downwards (as there is symmetry in time and height). So, during 10th second, the object was moving upwards, and during the 11th second, the object is coming downwards. Therefore, the distance S11 will be negative.

S11 ​= −(u–102.9)

Since they are equal, equating S10​ and S11​ :

u−93.1 = −u+102.9

2u = 196.0

u = 98 m/s

Explanation:  Hope it would help you...