If a ball is thrown straight up into the air with an initial velocity of 95 ft/s, it height in feet after t second is given by y=95t–16t2. Find the average velocity for the time period beginning when t=2 and lasting
i) 0.1 seconds
ii) 0.01 seconds
Nobody answered this question because it's impossible.
To find the average velocity for a given time period, we need to calculate the displacement (change in position) divided by the time interval.
In this case, we have the equation for the position of the ball as a function of time: y = 95t - 16t^2.
i) Average velocity for a time period of 0.1 seconds, starting from t = 2:
Step 1: Find the initial position at t = 2 seconds:
Plug in t = 2 into the equation: y(2) = 95(2) - 16(2^2) = 190 - 64 = 126 feet.
Step 2: Find the final position at t = 2.1 seconds (after 0.1 seconds):
Plug in t = 2.1 into the equation: y(2.1) = 95(2.1) - 16(2.1^2) = 199.5 - 74.16 = 125.34 feet.
Step 3: Calculate the displacement:
Displacement = Final position - Initial position = 125.34 - 126 = -0.66 feet.
Step 4: Calculate the average velocity:
Average velocity = Displacement / Time = -0.66 / 0.1 = -6.6 ft/s.
ii) Average velocity for a time period of 0.01 seconds, starting from t = 2:
Step 1: Find the initial position at t = 2 seconds (same as in part i):
Initial position = 126 feet.
Step 2: Find the final position at t = 2.01 seconds (after 0.01 seconds):
Plug in t = 2.01 into the equation: y(2.01) = 95(2.01) - 16(2.01^2) = 190.95 - 64.3216 = 126.6284 feet.
Step 3: Calculate the displacement:
Displacement = Final position - Initial position = 126.6284 - 126 = 0.6284 feet.
Step 4: Calculate the average velocity:
Average velocity = Displacement / Time = 0.6284 / 0.01 = 62.84 ft/s.