Answer:
Work is the energy transferred to or from an object via the application of force along a displacement.
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Answer:
<em>1,378.9ms²</em>
Explanation:
Given the following
Distance S = 70.6m
Time t = 0.32secs
Initial velocity = 0m/s
Required
Acceleration
Using the equation of motion
S = ut+1/2at²
Substitute
70.6 = 0+1/2a(0.32)²
70.6 = 0.0512a
a = 70.6/0.0512
a = 1,378.9
<em>Hence the acceleration is 1,378.9ms²</em>
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:
