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Answer:
5080.86m
Explanation:
We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:
We must consider that it's launched from the ground () and from rest (
), with an upwards acceleration
that lasts a time t=9.7s.
We calculate then the height achieved in part 1:
And the velocity achieved in part 1:
We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 () and its initial velocity is the one achieved in part 1 (
), now in free fall, which means with a downwards acceleration
. For the data we have it's faster to use the formula
, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:
Then, to get , we do:
And we substitute the values:
Answer:
(i) The angular speed of the small metal object is 25.133 rad/s
(ii) The linear speed of the small metal object is 7.54 m/s.
Explanation:
Given;
radius of the circular path, r = 30 cm = 0.3 m
number of revolutions, n = 20
time of motion, t = 5 s
(i) The angular speed of the small metal object is calculated as;
(ii) The linear speed of the small metal object is calculated as;