A ball is thrown straight down from the top of a 220-foot 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?

Answer 1

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

Answer 2

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.

Answer 3

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
0 = -16t^2 - 22t + 112
factor out a -2
0 = -2(8t^2 - 11t + 56)

use quadratic equation
(-11 ± √-1671)/16 = imaginary

use -2 to find v(t)

v = -32(2) - 22 = -86 ft/s

Answer 4

s(t) = -16t^2 - 22t + 220
s'(t) = ds/dt = -32t - 22
so at t=3 sec.
s'(3) = -32(3) - 22 = -118 ft/sec

when s(t) = 108 ft.
108 = -16t^2 - 22t + 220
16t^2 + 22t - 112 = 0
t = (22 + √7652)/-32 = -3.421115966 = 3.421 sec.

so at 108ft, the time is 3.421 sec.
use 3.421 sec. to find s'(t) at 108 ft.

s'(3.421) = -32(3.421) - 22 = -131.472 ft/sec (<- velocity at the height of 108ft)

Answer 5

where did t=3 come from

Answer 6

since it's asking for the velocity after it falls 108 feet, it isn't asking for when the position function is equal to 108, it wants it when it is 112 since it fell 108 of the 220 total feet leaving it at a y-level of 112. so you solve for "t" when the position function is equal to 112 and you get t=2. then you plug it into the velocity function to get -86 ft/s