Answer:
The height of the cliff is approximately 127.31 meters
Explanation:
The given parameters are;
The upward speed of the ball = 30 m/s
The time after the ball is thrown before it lands = 9 s
We have;
The time to maximum height;
v = u - g·t
Where;
v = The velocity at maximum height = 0 m/s
u = Initial velocity = 30 m/s
g = The acceleration due to gravity = 9.81 m/s²
∴ 0 = 30 - 9.81 × t
t = 30/9.81 = 3.06 s
The maximum height can be obtained from;
v² = u² - 2·g·s
v = 0 m/s
u² = 2·g·s
30² = 2 × 9.81 × s
s = 30²/(2×9.81) = 45.87 m
The time from maximum height to landing = 9 - 3.06 = 5.942 s
The height to landing
s = u·t + 1/2·g·t².
Here, u = 0 m/s
s = 0×5.942 + 1/2×9.81×5.942² = 173.1766 m
s = 173.1766 m.
The height of the cliff = 173.1766 m. - 45.87 m. ≈ 127.31 m.
