칼큘러스 james stewart 6th solution

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Note: the scale is distorted in this graph. For t = 0, the graph starts at (1, 0, 1). The upper loop shown
intersects the xz-plane at about (286751, 0, 286751).
(a) The shell’s trajectory is given by r(t) = 240√3 ti + (240t − 16t2 + 1600)j or x = 240√3 t,
y = 240t − 16t2 + 1600.
(b) Solving dy/dt = 240 − 32t = 0, we see that y is maximum when t = 15/2. The maximum
altitude is y(15/2) = 2500 ft.
(c) Solving y(t) = −16t2 + 240t + 1600 = −16(t − 20)(t + 5) = 0, we see that the shell hits the ground
when t = 20. The range of the shell is x(20) = 4800√3 ≈ 8314 ft.
(d) From (c), impact is when t = 20. The speed at impact is

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