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b)Findv(t), the velocity at time t.
c)Find s(t), the height at time t.
d)How high does it go?
Solution
Answer to part (a): The acceleration due to gravity is generally taken as -32 ft/sec2. Assuming effects such as wind resistance are negligable, a(t) = -32
I’ll define t=0 as the instant at which the rocket is launched.
v(t) is the integral of a(t), so v(t) = -32t +C. We know 100 = v(0) = -32*0 + C, so C = 100
Answer to part (b): v(t) = -32t + 100
s(t) is the integral of v(t). Therefore s(t) = -16t2 + 100t + D. We know 6 = s(0) = -16*02 + 100*0 + D, so D = 6.
Answer to part (c): s(t) = -16t2 + 100t + 6.
Maximum height is achieved when v(t) = 0, which occurs when t = 100/32 = 25/8. This tells us *when*. To find out how high, evaluate s(25/8).
Answer to part (d): s(25/8) = -16*(25/8)2 + 100*(25/8) + 6 = 162.25 ft