A ball is thrown vertically upward from the edge of a cliff with an initial speed of 18 m/s. It reaches the ground 5.0 s later. What is the height of the cliff above the ground? (Ignore air resistance and take g = 10 m/s².)

Correct answer

B. 35 m

Principle or equation

For uniformly accelerated motion, displacement is given by s = v0 t + (1/2) a t². Taking upward as positive, the acceleration is a = -g. The displacement from the throw point to the ground is negative (downward), so the cliff height is the magnitude of that displacement.

Why this answer is correct

Choose upward as positive. The initial velocity is v0 = +18 m/s, acceleration a = -10 m/s², and time t = 5.0 s. Displacement from the throw point is s = (18)(5) + 0.5(-10)(5²) = 90 - 125 = -35 m. The negative sign means the final position is 35 m below the throw point, so the cliff height is 35 m.

Example

If an object is thrown upward at 10 m/s from a cliff and lands 3 s later, s = 10(3) - 0.5(10)(9) = 30 - 45 = -15 m, so the cliff is 15 m high.