Calculus word problem help please?

Question

b)Findv(t), the velocity at time t.

c)Find s(t), the height at time t.

d)How high does it go?

Answer 1

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