I think the answer may be the letter B
It's average speed during that 26 seconds was about 4.77 m/s. Without seeing the graph, we can't tell if it was going faster or slower at any particular time during that period. All we can tell is its average for the full interval.
Answer:
The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.
Explanation:
Given that,
Initial velocity u= 128 ft/sec
Equation of height
....(I)
(a). We need to calculate the maximum height
Firstly we need to calculate the time

From equation (I)




Now, for maximum height
Put the value of t in equation (I)


(b). The number of seconds it takes the object to hit the ground.
We know that, when the object reaches ground the height becomes zero




Hence, The maximum height attained by the object and the number of seconds are 128 ft and 4 sec.
Answer:
2156J
Explanation:
Given parameters:
Height of lift = 10m
Mass = 22kg
Unknown:
Work done by the machine = ?
Solution:
Work done is the force applied to move a body through a certain distance.
So;
Work done = Force x distance
Here;
Work done = mass x acceleration due to gravity x height
Work done = 22 x 9.8 x 10 = 2156J