How much work does it take to accelerate a 9.0 kg object from rest to 27 m/s?
1 answer:
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The first part of the text is missing, you can find on google:
"A ball is thrown horizontally from the roof of a building 45 m. If it strikes the ground 56 m away, find the following values."
Let's now solve the different parts.
(a) 3.03 s
The time of flight can be found by analyzing the vertical motion only. The vertical displacement at time t is given by
where
h = 45 m is the initial height
g = 9.8 m/s^2 is the acceleration of gravity
When y=0, the ball reaches the ground, so the time taken for this to happen can be found by substituting y=0 and solving for the time:
(b) 18.5 m/s
For this part, we need to analyze the horizontal motion only, which is a uniform motion at constant speed.
The horizontal position is given by
where
is the horizontal speed, which is constant
t is the time
At t = 3.03 s (time of flight), we know that the horizontal position is x = 56 m. By substituting these numbers and solving for vx, we find the horizontal speed:
The ball was thrown horizontally: this means that its initial vertical speed was zero, so 18.5 m/s was also its initial overall speed.
(c) 35.0 m/s at 58.1 degrees below the horizontal
At the impact, we know that the horizontal speed is still the same:
we need to find the vertical velocity. This can be done by using the equation
where
is the initial vertical velocity
g is the acceleration of gravity
t is the time
Substituting t = 3.03 s, we find the vertical velocity at the time of impact:
So the magnitude of the velocity at the impact (so, the speed at the impact) is
The angle instead can be found as:
so, 58.1 degrees below the horizontal.