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

Answer:

27.52 meters is the height of the cliff.

Explanation:

Second equation of motion:

$h=u\times t+\frac{1}{2}g\times t^2$

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 = $9.8 m/s^2$

$h=0 m/s\times t+\frac{1}{2}g\times t^2$..[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 = $9.8 m/s^2$

$h=32 m/s\times (t-1.6)+\frac{1}{2}g(t-1.6)^2$..[2]

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

$0 m/s\times t+\frac{1}{2}g\times t^2=32 m/s\times (t-1.6)+\frac{1}{2}g(t-1.6)^2$

On solving for t:

t = 2.37 seconds

$h=0 m/s\times 2.37 s+\frac{1}{2}g\times t^2=\frac{1}{2}\times 9.8 m/s^2\times (2.37 s)^2$

h = 27.52 m

27.52 meters is the height of the cliff.

Answer 2

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