Answer:
A) 10.243 s
B) 514.64 m
C) 153.645 m
Explanation:
From projectile motion;
y(t) = h - ½gt²
Where;
h is the height of cliff
t is the time taken to fall from top of cliff down to half the height of the cliff
y(t) is height of the rock as function of time
We are told the rock falls half way of the cliff height.
Thus;
y = h/2
So;
h/2 = h - ½gt²
This gives;
h - h/2 = ½gt²
h/2 = ½gt²
h = gt² - - - - (1)
Also,the total free fall time from top of the cliff to ground would be;
T = t + 3
Thus, distance at this point is zero.
So;
0 = h - ½gT²
h = ½gT²
Putting t + 3 for T, we have;
h = (1/2)g(t + 3)² - - - 2
Inspecting eq 1 and eq 2,we have;
gt² = (1/2)g(t + 3)²
g will cancel out to give;
2t² = t² + 6t + 9
t² - 6t - 9 = 0
The roots are -1.243 or 7.243.
We will pick the positive value as time can't be negative.
Thus, T = 7.243 + 3 = 10.243 s
h = gt² = 9.81 × 7.243²
h = 514.64 m
Formula for horizontal distance is given by;
x = uT
Where u is initial speed given as;
u = 15 m/s
Thus;
x = 15 × 10.243
x = 153.645 m