Question

At t=0, a stone is dropped from a cliff above a lake; 1.6 s later another stone is thrown downward from the same point with an initial speed of 32 m/s. Both stones hit the water at the same instant. Find the height of the cliff.

Answer 1

Let equation A be
h=ut+12gt2 for 1st stone:

h=12 gt2

eq.-(A) for 2nd stone:

h=32m/s(t−1.6s)+12g(t−1.6s)2

h=32m/s∗t−51.5m+12gt2+12.544m−15.68m/s∗t

h=16.32m/s∗t+h−38.656m

t=2.3686s

now putting the value of t in eqn. A h=27.49m

Answer 2

27.52 meters is the height of the cliff.

Step-by-step explanation:

Second equation of motion:

`h=u* t+(1)/(2)g* t^2`

u = Initial velocity

t = Time taken to cover h distance

g = Acceleration due to gravity

Let the height of the cliff be h

Stone-1:

Initial velocity of the stone = u = 0 m/s

Time to cover h height of cliff, t = ?, g =
`9.8 m/s^2`

`h=0 m/s* t+(1)/(2)g* t^2`..[1] (Second equation of motion)

Stone-2:

Initial velocity of the stone = u' = 32 m/s,

Time to cover h height of cliff = t' = (t-1.6),

g =
`9.8 m/s^2`

`h=32 m/s* (t-1.6)+(1)/(2)g(t-1.6)^2`..[2]

Both stones are covering same distance or height of the cliff: [1]= [2]

`0 m/s* t+(1)/(2)g* t^2=32 m/s* (t-1.6)+(1)/(2)g(t-1.6)^2`

On solving for t:

t = 2.37 seconds

`h=0 m/s* 2.37 s+(1)/(2)g* t^2=(1)/(2)* 9.8 m/s^2* (2.37 s)^2`

h = 27.52 m

27.52 meters is the height of the cliff.