how long is the seashell in the air

Question

You drop a seashell into the from 40 feet. Write an equation that models the height in feet of the seashell above water after t seconds how long is the seashell in the air?

Answer 1

Y= distance of shell from the ground in feet

t= time in seconds

a= acceleration due to gravity =-16ft/sec^2

v(0)= initial velocity of the shell

y(0)= height from which shell was dropped.

Equation: 

Y=y(0)+v(0)*t+.5*a*t^2

at the instant the shell is dropped, the initial velocity is zero so the equation reduces to:

Y=40ft-.5*(-16ft/sec^3)*t^2 or,

Y=-16t^2+40

when the shell reaches the ground, the height (Y) will be zero, so:

0=40-16t^2

solving for t, gives

t=sqrt(40/16), or, sqrt(10)/2

Answer 2

This is a free-fall problem starting from rest.  Let up be positive, down be negative.

 

h₀ = initial height = 40 ft

v₀ = initial vertical speed = 0 ft/s (dropped from rest)

g = acceleration of gravity = 32 ft/s²

h(t) = height as a function of time

 

h(t) = h₀ + v₀t - (1/2)gt²

 

Plug in h₀ = 40 ft, v₀ = 0 ft/s, and g = 32 ft/s² and get

 

h(t) = 40 - 16t²

 

 

(h in ft, t in seconds)

 

When it reaches the surface, h = 0

 

0 = 40 - 16t²

16t² = 40

t = √(40/16) = 1.58 s

 

It is in the air 1.58 seconds before it reaches the surface.