Question

a particle at rest ,falls under gravity such that it travels 53.9m in last second of its journey .total time of fall is​

Answer 1

Answer:

The total time of fall is 7 sec.

explanation:

Given that,

Distance d = 53.9 m

Using equation of motion

S_{n}=u+\dfrac{1}{2}g{2n-1}S

n

=u+

2

1

g2n−1

Where, S = distance

u = initial velocity

n = time

Put the value in the equation

53.9=0+\dfrac{1}{2}\times9.8\times(2(t-1)-1)53.9=0+

2

1

×9.8×(2(t−1)−1)

53.9=4.9(2t-3)53.9=4.9(2t−3)

53.9+14.7=9.8t53.9+14.7=9.8t

t =\dfrac{53.9+14.7}{9.8}t=

9.8

53.9+14.7

t =7\ sect=7 sec

Hence, The total time of fall is 7 sec.