Exercise 2: Simulating a Ball Thrown Upward — Possible Solution
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GIVEN
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x0=0, v0=15 m/s, a=-9.8 m/s^2, dt=0.5, 2 steps (covering t=0 to
t=1)

STEP 1: EULER STEP 1 (t=0 to t=0.5)
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x_new = x + v*dt = 0 + 15*0.5 = 7.5
v_new = v + a*dt = 15 + (-9.8)*0.5 = 15 - 4.9 = 10.1

After step 1: x=7.5, v=10.1

STEP 2: EULER STEP 2 (t=0.5 to t=1.0)
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x_new = x + v*dt = 7.5 + 10.1*0.5 = 7.5 + 5.05 = 12.55
v_new = v + a*dt = 10.1 + (-9.8)*0.5 = 10.1 - 4.9 = 5.2

After step 2: x=12.55, v=5.2

STEP 3: COMPARE TO THE EXACT FORMULA AT t=1
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x(1) = x0 + v0*t + 0.5*a*t^2 = 0 + 15*1 + 0.5*(-9.8)*1^2
= 15 - 4.9 = 10.1

RESULT
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Euler's method (2 steps, dt=0.5): x=12.55
Exact analytical solution: x=10.1

Error: |12.55 - 10.1| = 2.45

This is a noticeably larger error than this chapter's own worked
example, which used a much finer dt=0.01 - directly consistent with
this chapter's own verified finding that Euler's method's error
shrinks linearly with dt. A large, coarse dt=0.5 here produces a
correspondingly large error.

WHY THIS WORKS AS AN ANSWER
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Both Euler steps are computed explicitly showing the position and
velocity update separately at each step (using the OLD velocity for
the position update, matching this chapter's own method exactly),
and the resulting error is explicitly connected back to this
chapter's own dt-vs-error finding to explain why it's larger than
the chapter's own more finely-stepped example, rather than treated
as an unexplained discrepancy.