Can you explain how to figure this out?
A helicoptor accelerates at a rate of 2m/s^2. If you drop a rock out of the helicoptor 3 seconds after takeoff how high is the helicoptor when the rock hits the ground.
Thanks
To figure out the height of the helicopter when the rock hits the ground, we need to use a few formulas of motion.
First, let's determine the time it takes for the rock to hit the ground. We know that the rock is dropped 3 seconds after takeoff, which means it has been in the air for a total time of (3 + t) seconds, where "t" is the time it takes for the rock to hit the ground.
Next, we'll find the vertical distance traveled by the rock using the equation of motion:
s = ut + (1/2)at^2
Where:
s = vertical distance traveled by the rock
u = initial velocity (in this case, 0 m/s since the rock is dropped)
a = acceleration due to gravity (which is approximately 9.8 m/s^2)
t = time
Since the rock falls due to gravity, the acceleration is negative (-9.8 m/s^2).
Now, let's calculate the vertical distance traveled by the rock using the equation of motion:
s = 0t + (1/2)(-9.8)t^2
s = -4.9t^2
The negative sign indicates that the distance is in the opposite direction of the initial position (up from the ground).
Now, we can substitute the total time, (3 + t), into the equation:
s = -4.9(3 + t)^2
Finally, to find the height of the helicopter when the rock hits the ground, we need to consider that the rock is dropped from the helicopter. So, the height of the helicopter is given by subtracting the distance traveled by the rock from the total height covered by the helicopter from the point of takeoff. Assuming the initial height of the helicopter is zero:
Height of the helicopter = Total height covered by the helicopter - distance traveled by the rock
Height of the helicopter = 0 - s
Substituting the value of "s" into the equation:
Height of the helicopter = 0 - (-4.9(3 + t)^2)
This equation will give you the height of the helicopter when the rock hits the ground.