Under the influence of a complicated collection of physical fields, the altitude h of a rocket from its moment at launch until its fall to Earth is given by the formula h(t)= t^4 - 8t³ + 375 where the altitude h is measured in feet and the time launch t is measured in seconds. What is the initial altitude? When does the projectile impact Earth? What is the max. altitude? What is its max upward velocity??? What is its max downward velocity??? What is its max upward acceleration??? What is its max downward acceleration???
2006-11-10 14:01:38 · 4 answers · asked by hippiechic2241 1 in Science & Mathematics ➔ Physics
h(t)= t^4 - 8t³ + 375
h(0) = 375 ft
time of impact: It ain't gonna hit with this equation!
t^4 - 8t³ + 375 = 0
v = 4t³ - 24t²
a = 12t² - 48t
da/dt = 24t - 48
24t = 48
t = 2
a = 12*4 - 48*2 = -48 ft/sec^2 @ t = 2
2006-11-10 14:33:16 · answer #1 · answered by Helmut 7 · 1⤊ 0⤋
Initial altitude is 375 feet. Set t = 0 and solve.
Max downward acceleration will be 32 ft/sec or 1 g.
Maximum height can be found by differentiating the height formula and solving for t. 4t^3 -24t^2 = 0 (t=6) Then put the value for t in the original height equation.
The rest of the material is easily solvable.
2006-11-10 14:14:04 · answer #2 · answered by eriurana 3 · 0⤊ 0⤋
no sure sorry
2006-11-10 14:05:17 · answer #3 · answered by twisted 3 · 0⤊ 0⤋
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2006-11-10 14:04:28 · answer #4 · answered by B 3 · 0⤊ 0⤋