Find the initial velocity of an object if the velocity after 2 seconds is 22ft/s.

Question

If an object is shot upward with an initial velocity, v0v0, in feet per second (ft/s), the velocity, v, in ft/s is given by the formula v=v0−32tv=v0−32t, where t is time in seconds. Find the initial velocity of an object if the velocity after 22 seconds is 22ft/s22⁢ft/s.

Answer 1

h(t) = -16t^2 + vot + ho

where ho = initial height measured in feet

, vo = initial velocity measured in feet per second,

h(t) = height measured in feet, at time t measured in seconds

take the derivative of h(t) to get velocity at time t

v(t) = h'(t) =-32t +vo

v(22) = -32(22) + vo = 22

vo = 22+32(22) = 726 ft/sec = initial velocity

the object reaches maximum height when v(t)=0 in 22.6875 seconds

in about 22.7 seconds

in 22 seconds it's slowing down and close to maximum height

in 22 seconds it's at h=7,986 feet

in another 7/10th of a second it reaches max height of 8,235.525

(plus ho if it didn't start at ground level)

it takes 45 3/8 seconds to fall back to the ground, twice the time to reach max height

that's the solution the way you've written the problem

but your use of "22" both t and v(22) is a little suspicious, as if maybe the problem got miscopied or had a mistake in it from the beginning