A man releases a stone ar the top edge of a tower during the last second of its travel the stone falls through a distance of (9/
2 answers:
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Explanation:
Given that,
Distance, s = 47 m
Time taken, t = 8.6 s
Final speed of the truck, v = 2.3 m/s
Let u is the initial speed of the truck and a is its acceleration such that :
.............(1)
Now, the second equation of motion is :
Put the value of a in above equation as :
u = 8.63 m/s
So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.
The initial height of the first body is given by:
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.
where
g is the gravitational acceleration
t is the time it takes for the body to reach the ground
Substituting t=1 s, we find
The second body takes takes t=2 s to reach the ground, so it was located at an initial height of
The second body started its fall 1 second before the first body, therefore when the second body started its fall, the first body was located at its initial height, i.e. at 4.9 m from the ground.