Question

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 . The acceleration period lasts for time 5.00 s until the fuel is exhausted. After that, the rocket is in free fall. Find the maximum height ymax reached by the rocket. Ignore air resistance and assume a constant free-fall acceleration equal to 9.80 m/s2 .

Answer 1

Hi there!

We can begin by using the following formulas:

Δd = vit + 1/2at²

vf² = vi² + 2ad

v = at

We can break this problem into 2 parts:

Distance traveled with acceleration:

Δd = vi(t) + 1/2at²

The rocket starts from rest so disregard the "vi(t)":

Δd = 1/2at²

Plug in the given acceleration and time:

Δd = 1/2(29.4)(5)²

Δd = 367.5 m

Distance traveled after thrust stops:

Find the displacement up to the top of the trajectory using the equation:

vf² = vi² + 2ad, where vf = 0 m/s

Calculate the initial velocity using the equation v = at:

v = (29.4)(5) = 147 m/s

Use the above equation and plug in the given values:

0² = (147)² + 2(-9.8)d

Solve for d:

0 = 21,609 - 19.6d

19.6d = 21,609

d = 1,102.5 m

Add the two displacements:

d1 + d2 = 1,470 m