let the height of the person with marshmallow on her head be "h"
consider the motion of the marshmallow after it is dropped from bridge.
Y₀ = initial position of the marshmallow above the ground = 5.71 m
Y = final position of marshmallow on head of person = h
v₀ = initial velocity of the marshmallow = 0 m/s
a = acceleration due to gravity = - 9.8 m/s²
t = time of travel for marshmallow = 0.921 sec
Using the kinematics equation
Y = Y₀ + v₀ t + (0.5) a t²
inserting the values
h = 5.71 + 0 (0.921) + (0.5) (-9.8) (0.921)²
h = 5.71 - 4.16
h = 1.55 m
Answer:
Part a)
T = 3.96 s
Part b)
T = 1.98 s
Part c)
T = 2.8 s
Explanation:
As we know that time period of spring block system is given as

T = 2.8 s
Part a)
If the mass of the block attached is doubled
then we will have



Part b)
If the spring constant is doubled
then we have



Part c)
If the amplitude is halved but mass and spring constant will remain the same
so here we know that time period does not depends on Amplitude
so we will have

T = 2.8 s
The substance has changed, It dissolved, and it evaproted.
Answer:
the answer will be 3.763 seconds