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 height= h and at time t first stone touch the water.(g=9.8m/s^2)then eqn. of motion.
\[h=ut+\frac{ 1 }{ 2 }g t^2\]
for 1st stone:\[h=\frac{ 1 }{ 2 }\ g t^2\]eq.-(A)


for 2nd stone:
\[h=32m/s(t-1.6s)+\frac{ 1 }{ 2 } g(t-1.6s)^2\]\[h=32m/s*t-51.5m+\frac{ 1 }{ 2 } g t^2 +12.544m-15.68m/s*t\]
\[h=16.32m/s*t + h - 38.656m\]
\[t=2.3686s\]
now putting the value of t in eqn. A
\[h=27.49m\]