Question

A ball is thrown into the air with a velocity of 34 ft/s. It's height in feet after t seconds is given by y=34t-26(t)^2. Find the avg velocity for the time period beginning when t(0)= 3 seconds and lasting for the given time t= .01 seconds: ???

Answer 1

Answer:

The average velocity of the ball at the given time interval is -122.3 ft/s

Step-by-step explanation:

Given;

velocity of the ball, v = 34 ft/s

height of the ball, y = 34t - 26t²

initial time, t₀ = 3 seconds

final time, t = 3 + 0.01 = 3.01 seconds

At t = 3 s

y(3) = 34(3) - 26(3)² = -132

The average velocity of the ball in ft/s is given as;

$V_{avg} = \frac{y(3.01)-y(3)}{3.01 -3}\\\\V_{avg} = \frac{34(3.01)-26(3.01)^2-y(3)}{3.01 -3}\\\\V_{avg} = \frac{-133.223-y(3)}{0.01}\\\\V_{avg} = \frac{-133.223-(-132)}{0.01}\\\\V_{avg} =\frac{-1.223}{0.01}\\\\V_{avg} = -122.3 \ ft/s$

Therefore, the average velocity of the ball at the given time interval is -122.3 ft/s