Answer:
Vf = 3.67 [m/s]
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
Vf = final velocity [m/s]
Vi = initial velocity = 4.3 [m/s]
a = acceleration or desacceleration = 0.5 [m/s²]
x = distance = 5 [m]
Note: The negative sign in the above equation means that the velocity of the ball is decreasing (desacceleration).
Now replacing:
Vf² = (4.3)² - (2*0.5*5)
Vf² = 18.49 - 5
Vf² = 13.49
using the square root, we have.
Vf = 3.67 [m/s]
The average velocity for the time period beginning when t=1 and lasting
(i) 0.01 seconds = 63.84 ft/s
(ii) 0.001 seconds = 63.984 ft/s
Given that a ball is thrown with an initial velocity = 80 ft/s
Let 'y' be the height in feet after 't' seconds.
Given, gives the height in 't' seconds.
Average velocity = Rate of change of distance
= Change in distance/Change in time.
The initial time can be taken as 0 s.
When t =1 s, y = 80 - 16 = 64 ft
(1) t = 0.01 s
y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft
Average velocity = (64 - 0.7984) / (1 -0.01) = 63.84 ft/s
(2) t = 0.001 s
y = 80 x 0.001 - 16 x 0.001 x 0.001 = 0.079984 ft
Average velocity = (64 - 0.079984) / (1 -0.001) = 63.984 ft/s
The question is incomplete. Find out the complete question below:
If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by .Find the average velocity for the time period beginning when t=1 and lasting
(i) 0.01 seconds
(ii) 0.001 seconds
Learn more about average velocity at brainly.com/question/6504879
#SPJ4