Calculus
posted by Susana on .
A ball is thrown straight down from the top of a 220foot building with an initial velocity of 22 feet per second. What is its velocity after 3 seconds? What is its velocity after falling 108 feet?

equation:
s = 16t^2  22t + 220
v = ds/dt = 32t  22
so at t=3
v = 16(3)  22 = 70 ft/sec
when s = 108
108 = 16t^2  22t + 220
16t^2 + 22t  112 = 0
t = (22 ± √7652)/32 = 2.046 sec or a negative time
so when t = 2.046
v = 16(2.046)  22 =  54 ft/s 
s = 16t^2  22t + 220
v = ds/dt = 32t  22
so at t=3
v = 32(3)  22 = 118 ft/sec
when s = 108
108 = 16t^2  22t + 220
16t^2 + 22t  112 = 0
t = (22 ± √7652)/32 = 2.046
so when t = 2.046
v = 32(2.046)  22 = 87.472 ft/s
The process is correct but the t=2.046 should be substituted to the first derivative which is velocity function.