Amodel rocket is fired vertically from the ground with a constant acceleration of 45.3 m/s2 for 1.46 s at which time its fuel is exhausted. what is the maximum height (in m) reached by the rocket?

271 m

Explanation:

We can divide the motion into two stages.  In the first stage, the rocket is accelerating upward.  In the second stage, the fuel has exhausted, and the rocket is in free fall.

For the first stage, we are given:

y₀ = 0 m

v₀ = 0 m/s

a = 45.3 m/s²

t = 1.46 s

Find: y, v

y = y₀ + v₀ t + ½ at²

y = (0) + (0) (1.46) + ½ (45.3) (1.46)²

y = 48.3 m

v = at + v₀

v = (45.3) (1.46) + (0)

v = 66.1 m/s

In the second stage, we are given:

y₀ = 48.3 m

v₀ = 66.1 m/s

v = 0 m/s

a = -9.8 m/s²

Find: y

v² = v₀² + 2a(y - y₀)

(0)² = (66.1)² + 2(-9.8) (y - 48.3)

y = 271 m

The rocket reaches a maximum height of 271 m.

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