A ball is thrown upward from a cliff. The ball rises to its maximum height and then falls into the valley below. Suppose the ball is thrown upward at 30m/s and lands 9s after it is thrown. What is the height of the cliff?

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.

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