3. A student throws a calculator upward at 22 m/s. How far from the ground will the calculator be when it has half the speed it was thrown with? How much time did it take to reach this height?

18 m

1.1 s

Explanation:

Given:

v₀ = 22 m/s

v = 11 m/s

a = -10 m/s²

Find: Δy

v² = v₀² + 2aΔy

(11 m/s)² = (22 m/s)² + 2 (-10 m/s²) Δy

Δy ≈ 18 m

Find: t

v = at + v₀

11 m/s = (-10 m/s²) t + 22 m/s

t = 1.1 s

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