Question

If a ball is thrown straight up into the air with an initial velocity of 40 ft/s, it height in feet after t second is given by y=40t−16t2. Find the average velocity for the time period begining when t=2 and lasting

(i) 0.5 seconds

(ii) 0.1 seconds

(iii) 0.01 seconds

Finally based on the above results, guess what the instantaneous velocity of the ball is when t=2.

Answer 1

To find the average velocity, use the formula for change in position divided by change in time. Based on the average velocities, estimate the instantaneous velocity when t=2.

Step-by-step explanation:

To find the average velocity for the time period beginning when t=2 and lasting:

• (i) 0.5 seconds: We can use the formula for average velocity, which is change in position divided by change in time. In this case, the change in position is y(2.5) - y(2) and the change in time is 0.5 seconds. Plug in the values and calculate.
• (ii) 0.1 seconds: Follow the same steps as in (i) but with a change in time of 0.1 seconds.
• (iii) 0.01 seconds: Use the same formula but with a change in time of 0.01 seconds.
• Based on the above results, we can make an estimate for the instantaneous velocity when t=2 by looking at the average velocities as the time interval gets smaller. As the time interval approaches zero, the average velocity becomes closer to the instantaneous velocity.

Answer 2

We should first calculate the highest point that ball reaches.
y' = 40 - 32t = 0
t = 40/32 = 1,25s
y = 25 feet.

To calculate average velocity we use simple formula:
Vav=s/t where s is traveled distance by the time t.



for t=2 we calculate y
y = 16
(i) for t = 2,5 y = 0
Vav = 16/0.5 = 32 feet/s
(ii) for t = 2.1 y = 13.44
Vav = 2.56/0.1 = 25.6 feet/s
(iii) for t = 2.01 y = 15.7584
Vav = 0.2416/0.01= 24.16 feet/s

it seems like the answer would be 24 feet/s. There is a way to calculate that.