Supposing there's no air
resistance, horizontal velocity is constant, which makes it very easy to solve
for the amount of time that the rock was in the air.
Initial horizontal
velocity is: <span>
cos(30 degrees) * 12m/s = 10.3923m/s
15.5m / 10.3923m/s = 1.49s
So the rock was in the air for 1.49 seconds. </span>
<span>
Now that we know that, we can use the following kinematics
equation:
d = v i * t + 1/2 * a * t^2
Where d is the difference in y position, t is the time that
the rock was in the air, and a is the vertical acceleration: -9.80m/s^2. </span>
<span>
Initial vertical velocity is sin(30 degrees) * 12m/s = 6 m/s
So:
d = 6 * 1.49 + (1/2) * (-9.80) * (1.49)^2
d = 8.94 + -10.89</span>
d = -1.95<span>
<span>This means that the initial y position is 1.95 m higher than
where the rock lands. </span></span>
When they meet the 40-kg boy would have moved a distance of 6 m.
<h3>Distance moved by the 40 kg boy</h3>
Apply the principle of center mass;
Take the 40 kg mass as the reference point;
X(40 kg) = (40kg x 0 + 60kg x 10 m)/(40 kg + 60 kg)
X(40 kg) = (600)/(100)
X(40 kg) = 6 m
Thus, when they meet the 40-kg boy would have moved a distance of 6 m.
Learn more about distance here: brainly.com/question/2854969
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The dummy's acceleration is 11 m/s^2
(also known as 11 meters per sec. per sec.)
a = F/m
= 825 N/75 kg
= 11 m/s^2
The answer should be 2times3.14 which equals 6.28