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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 39.2 m/s^2. The acceleration period lasts for time 10.0 ss until the fuel is exhausted. After that, the rocket is in free fall.

Required:
Find the maximum height ymax reached by the rocket.

Respuesta :

MathPhys

Answer:

9800 m

Explanation:

During acceleration, given:

v₀ = 0 m/s

a = 39.2 m/s²

t = 10.0 s

Find: v and Δy

v = at + v₀

v = (39.2 m/s²) (10.0 s) + 0 m/s

v = 392 m/s

Δy = v₀ t + ½ at²

Δy = (0 m/s) (10.0 s) + ½ (39.2 m/s²) (10.0 s)²

Δy = 1960 m

During free fall, given:

v₀ = 392 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (392 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 7840 m

Therefore, h = 1960 m + 7840 m = 9800 m.