Answer:
a) t = 3.771s
b) h = 69.68 m
Explanation:
Given;
total distance covered = h
Using the second equation of motion;
h = ut + 0.5at² .....1
We know u=0, since the object fall from rest.
and also acceleration a = g (acceleration due to gravity)
The equation 1 becomes;
h = 0.5gt² .......2
Equation 2 is the overall equation of motion of the object.
Where;
h = height of fall
t = total time of fall
The motion have two phases.
1) distance travelled in the first phase is
h₁ = h - 0.46h = 0.54h
Time taken for the first phase is;
t₁ = t-1
2) distance travelled in the second phase is
h₂ = 0.46h
Time taken for second phase is
t₂ = 1.00s
Writing the equation of motion of the first phase where the object starts from rest, we have;
0.54h = 0.5g(t-1)² .....3
Substituting equation 2 into 3, we have;
0.54(0.5gt²) = 0.5g(t-1)²
divide both sides by 0.5g
0.54t² = (t-1)²
(t-1)² - 0.54t² = 0
Simplifying the equation above, we have;
t² - 2t +1 - 0.54t² = 0
Then we have a quadratic equation;
0.46t² - 2t +1 = 0
solving the quadratic equation, we have
t = 3.771 or t = 0.5764
Since t cannot be less than 1,
Recall that t₁ = t-1, and t₁ cannot be negative. so t>1
t = 3.771s
b) Using equation 2
h = 0.5gt²
h = 0.5×9.8×3.771²
h = 69.68 m
