Answer:
44.3 m/s
Explanation:
Given that a ball is thrown horizontally from the top of a building 100m high. The ball strikes the ground at a point 120 m horizontally away from and below the point of release.
What is the magnitude of its velocity just before it strikes the ground ?
The parameters given are:
Height H = 100m
Since the ball is thrown from a top of a building, initial velocity U = 0
Let g = 9.8m/s^2
Using third equation of motion
V^2 = U^2 + 2gH
Substitute all the parameters into the formula
V^2 = 2 × 9.8 × 100
V^2 = 200 × 9.8
V^2 = 1960
V = 44.27 m/s
Therefore, the magnitude of its velocity just before it strikes the ground is 44.3 m/s approximately
Choice c it's confusing to understand but it's c
r = 30m
v = 2.5 m/s
m = .02kg
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v = r*omega
omega = (delta thaeda)/(delta time)
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2.5 m/s = 30m*omega
.08333... = omega
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*note, omega stands for angular velocity
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Also, it's been a while since I've done a physics problem like this so I could be wrong. I think this is the answer though :)
