Derivative Calculus motion questions?

Question

Ball thrown down from 220 ft building with initial velocity of -22 ft/s
What is its initial velocity after 3 seconds. What is its velocity after falling 108 ft
Position function s(t)= -16t^2 +Vot +So
vo=initial velocity
The part for what is the initial velocity after 3 seconds i got. i just took the derivative of the position function. I don't understand how to do the second part Someone please show me the steps to find velocity after falling 108ft

Answer 1

the 'answer' to the position function is the position, or height.
so set s(t) = height = 112 (220ft minus the 'after falling 108ft')
solve for t to know at what time the ball passes that point. then plug that t into the velocity function.

Answer 2

a) properly we could prefer to correctly known how lots time it took to bypass from factor O to show A. It tells us that it got here to sit down down back on A so as meaning 30t - 3t^2 has yet another root as properly 0. 0 = -3t^2 + 30t <==> means that t = 0, 10 so it takes 10 s to get to show A displacement, S, is only the fundamental of velocity so S = 15t^2 - t^3 evaluate S(10) - S(0) and u get S = 500 m b) Acceleration is only the by-manufactured from velocity so A = 30 - 6t evaluate it at 10 seconds. so A (10) = -30 m/s^2 c) trouble-free calculus optimization question. get the by-manufactured from velocity, this is acceleration, and set it equivalent to 0. sparkling up for t. 0 = 30 - 6t t = 5s now, evaluate the speed at 5 s V (5) = seventy 5 m/s