Question

An object is projected vertically upwards . if distance covered in 4th second and 7th second are equal.find time of flight

Answer 1

Answer:

11 is the answer of your question

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Answer 2

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 5th second:

(put a = -9.8 and n = 5)

S5 =u–44.1

Distance covered in the 6th second:

(put n =6, a=-9.8 in equation):

S6 = -(u–53.9)

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 5th second, the object was moving upwards, and during the 6th second, the object is coming downwards. Therefore, the distance S6 will be

negative.

S6 =−(u–53.9)

Since they are equal, equating S5

and S6

:

u−44.1=−u+53.9

2u=98

u=49m/s

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