University Physics I - Classical Mechanics, 2019

2.5. EXAMPLES 47
which the rocket reaches the top of its trajectory, and then starts coming down. The corresponding
displacement is, by Eq. (2.16),
so the maximum height it reaches is 30.4 m.
Δy top = v i2 (t top − t i2 ) − 1 2 g(t top − t i2 ) 2 =20.4m
At the end of the full 3-second interval, the rocket’s displacement is
Δy 2 = v i2 Δt 2 − 1 2 g(Δt 2) 2 =15.9m
(so its height is 25.9 m above the ground), and the final velocity is
v f2 = v i2 − gΔt 2 = −9.43 m s .
(c) The acceleration for this part is (v f3 − v i3 )/Δt 3 =(−5+9.43)/1 =4.43 m/s 2 .Notethepositive
sign. The displacement is
so the final height is 25.9 − 7.21 = 18.7 m.
Δy 3 = −9.43 × 1+ 1 2 × 4.43 × 12 = −7.22 m
(d) This is just motion with constant speed to cover 18.7 m at 5 m/s. The time it takes is 3.74 s.
The graphs for this motion are shown earlier in the chapter, in Figure 2.3.