Question

A rock is dropped from the top of a tall building. The rock's displacement in the last second before it hits the ground is 46 % of the entire distance it falls. Find the height

Answer 1

Answer:

height is 69.68 m

Explanation:

given data

before it hits the ground =  46 % of entire distance

to find out

the height

solution

we know here acceleration and displacement that is

d = (0.5)gt²     ..............1

here d is distance and g is  acceleration and t is time

so when object falling it will be

h = 4.9 t²   ....................2

and in 1st part of question

we have (100% - 46% ) = 54 %

so falling objects will be there

0.54 h = 4.9 (t-1)²       ...................3

so

now we have 2 equation with unknown

we equate both equation

1st equation already solve for h

substitute h in the second equation and find t

0.54 × 4.9 t² = 4.9 (t-1)²  

t = 0.576 s and  3.771 s

we use here 3.771 s because  0.576 s is useless displacement in the last second before it hits the ground is 46 % of the entire distance it falls

so take t = 3.771 s

then h from equation 2

h = 4.9 t²

h = 4.9 (3.771)²

h =  69.68 m

so height is 69.68 m