Answer:
a)Height of the cliff = 6.19 m
b) t = 0.571 s
Explanation:
We need to first calculate the time it takes the rock to go up and come down to the cliff's level first.
Using the equations of motion,
g = -9.8 m/s², u₁ = Initial velocity = 8.03 m/s,
v = velocity at maximum height = 0 m/s
t₁ = time to reach maximum height = ?
a) v₁ = u₁ + gt₁
0 = 8.03 - 9.8t
t₁ = 8.03/9.8 = 0.819 s
Total time of flight for the rock to leave the cliff and reach maximum height = 0.819 s
Height that the rock attains, measured from the top of the cliff to the maximum height, y₁ = ?
y₁ = ut₁ + gt₁²/2 = 8.03(0.819) - 9.8(0.819²)/2 = 3.29m
Total time of flight of the rock, t = t₁ + t₂
Subtracting the time obtained from the first part (0.819 s) from the total time would give the time it took the rock to travel from maximum height reached to the ground.
t₂ = 2.21 - 0.819 = 1.391 s
Let y = total height fallen through = y₁ + (height of the cliff) = y₁ + H = ?
g = 9.8 m/s²
t₂ = 1.391 s
u₂ = initial velocity for 2nd phase of rock motion = velocity at maximum height = 0 m/s
y = u₂t₂ + gt₂²/2
y = 0 + 9.8(1.391²)/2
y = 9.48 m
y = y₁ + H
9.48 = 3.29 + H
H = 9.48 - 3.29 = 6.19 m
b) y = H = 6.19 m, u = 8.03 m/s, t = ?,
y = ut + gt²/2
6.19 = 8.03t + 9.8t²/2
4.9t² + 8.03t - 6.19 = 0
Solving the quadratic eqn
t = -2.21 s or 0.571 s
Since time can only be positive,
t = 0.571 s
(PROVED)
