A marshmallow is dropped from a 5.71 meter high pedestrian bridge, and 0.921 seconds later, it lands right on the head of an unsuspecting person walking underneath. How tall is the person with the marshmallow on her head?

a: 1.75 m

b: 1.70 m

c: 1.66 m

d: 1.51 m

e: 1.55 m

f: 1.89 m

g: 1.49 m

h: 1.62 m

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