Question

a body is thrown vertically upwards with initial velocity 'u' reaches maximum height in 6 seconds. What is the ratio of distances travelled by the body in the first second and the seventh second?

Answer 1

Answer:

The initial velocity = u

Final velocity v = 0

Time taken = 6 secs

v = u + at

0 = u + 6a

As the ball is thrown up, the acceleration a = -g = -9.8

0 = $u - 9.8\times 6$

u = 58.8 m/s

formula to find the distance covered in nth second $S_n\quad =\quad u+(\frac { a }{ 2 } )(2n-1)$

for n =1,

$S_1\quad =\quad 58.8-(\frac { 9.8 }{ 2 } )(1)= 58.8 - 4.9 = 53.9 m$

For n = 7

$S_7\quad =\quad 58.8-(\frac { 9.8 }{ 2 } )(13)= -4.9m$

The negative sign indicates, the body is moving in downward direction.

Ratio for the distance traveled between 1st and 7th second = $\frac { 53.9 }{ 4.9 } = 11:1$