A ball is dropped from the top of the building.it initially moves at 4.0 m/s after 0.5 seconds it moves at 3.8m/s what force is
1 answer:
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Answer:
∆h = 0.071 m
Explanation:
I rename angle (θ) = angle(α)
First we are going to write two important equations to solve this problem :
Vy(t) and y(t)
We start by decomposing the speed in the direction ''y''
Vy in this problem will follow this equation =
where g is the gravity acceleration
This is equation (1)
For Y(t) :
We suppose yi = 0
This is equation (2)
We need the time in which Vy = 0 m/s so we use (1)
So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)
2.236 m is the maximum height from the shell (in which Vy=0 m/s)
Let's calculate now the height for t = 0.555 s
The height asked is
∆h = 2.236 m - 2.165 m = 0.071 m
