A stone is thrown vertically upward with an initial speed of 15 m/s from the edge of a cliff. It falls past the edge and hits the ground 4.0 s after being thrown. What is the height of the cliff above the ground? (Take g = 10 m/s^2 and ignore air resistance.)

Correct answer

A. 20 m

Principle or equation

For uniformly accelerated motion, displacement is given by s = v0 t + (1/2) a t^2, where upward is taken as positive and a = -g.

Why this answer is correct

Choose upward as positive. The displacement from the throwing point to the ground is s = v0 t + (1/2)(-g)t^2 = (15)(4.0) - (1/2)(10)(4.0)^2 = 60 - 80 = -20 m. The negative sign means the final position is 20 m below the starting point, so the cliff is 20 m high.

Example

If an object is thrown upward at 10 m/s and returns to the same level after 2 s, its displacement is zero: 10×2 - 0.5×10×4 = 0.