Final answer:
The velocity of the ball after 2 seconds is -86 ft/s. The velocity of the ball after falling 344 feet is -170.48 ft/s.
Step-by-step explanation:
To find the velocity of the ball after 2 seconds, we need to substitute the value of t = 2 into the velocity function. The velocity function is given as v(t) = -32t + vo, where vo is the initial velocity. Given that the initial velocity is -22 ft/s, we can calculate the velocity after 2 seconds using v(2) = -32(2) + (-22) = -64 -22 = -86 ft/s.
To find the velocity after falling 344 feet, we need to find the time it takes to fall 344 feet and then substitute this time into the velocity function. The equation for the distance fallen is given as s(t) = -16t^2 + vot + so, where so is the initial position. Given that the initial position is 425 feet, we can set s(t) = 425 - 344 = 81 and solve for t. The equation becomes 81 = -16t^2 - 22t + 425. By solving this quadratic equation, we find two possible values for t: t = -5.64 s and t = 4.64 s. Since time cannot be negative in this case, we can disregard the negative value. Thus, the velocity after falling 344 feet is v(4.64) = -32(4.64) + (-22) = -148.48 - 22 = -170.48 ft/s.