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3 years ago

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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 . Th

e 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 .

1 answer:

gtnhenbr [62]3 years ago

8 0

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

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By using Suvat equation:

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The complete question is:

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?

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