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Answer:
The ball is dropped at a height of 9.71 m above the top of the window.
Explanation:
<u>Given:</u>
- Height of the window=1.5 m
- Time taken by ball to cover the window height=0.15
Now using equation of motion in one dimension we have

Let u be the velocity of the ball when it reaches the top of the window
then

Now u is the final velocity of the ball with respect to the top of the building
so let t be the time taken for it to reach the top of the window with this velocity

Let h be the height above the top of the window

The distance it traveled and the time that it took to travel that distance
Answer:
Explanation:
The spring is stretched by .5 m and then released that means its amplitude of oscillation A is 0.5 m .
A = 0.5 m
After the release at one extreme point , the mass comes to rest again at another extreme point after half the time period ie
T / 2 = .3 s
T = 0.6 s
Angular velocity
ω = 
ω = 
ω = 10.45
Maximum velocity = ω A
ω and A are angular velocity and amplitude of oscillation.
Maximum velocity = 10.45 x .5
= 5.23 m /s