Respuesta :
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
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:
27.52 meters is the height of the cliff.
Explanation:
Second equation of motion:
[tex]h=u\times t+\frac{1}{2}g\times t^2[/tex]
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 = [tex]9.8 m/s^2[/tex]
[tex]h=0 m/s\times t+\frac{1}{2}g\times t^2[/tex]..[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 = [tex]9.8 m/s^2[/tex]
[tex]h=32 m/s\times (t-1.6)+\frac{1}{2}g(t-1.6)^2[/tex]..[2]
Both stones are covering same distance or height of the cliff: [1]= [2]
[tex]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[/tex]
On solving for t:
t = 2.37 seconds
[tex]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[/tex]
h = 27.52 m
27.52 meters is the height of the cliff.