Does the initial velocity of an object (being dropped from a certain height) affect the observed acceleration?
We did a lab in my AP Physics class, where we dropped an object next to some measuring device, that could tell how fast the object was dropping and create a graph of position (on the ruler that we dropped) vs. time. The question also asks, "Would throwing the [object] downward, but letting go before it enters the [measuring thing], change the observed acceleration?"
- oldprofLv 71 decade agoFavourite answer
If all this happens in a vacuum, no. There would be no change because g = GM/r^2, the free fall acceleration in a vacuum is affected only by the source mass M and the distance r from that mass. Earth remains as the mass and the distance remains as the radius of Earth wherever you're doing the experiment. So g remains the same no matter what speed the falling object starts out at.
But, this is important, if you are doing the experiment in air, then, yes, there would be a measurable difference.
That results because, in air, drag changes the acceleration from g to a = g (1 - D/W) < g something less than g as the velocity of the object increases. The drag varies with the square of the falling velocity V, e.g., D ~ V^2, and W = mg is the weight of the object with mass m. So when the initial velocity is U = V > 0 because the object is tossed, there will be drag right away. And the acceleration will be less than g right away.
Note that U = A t = (F/m) t is the initial velocity of the toss of t seconds with F force on the object with m mass. A = 0 after the object leaves the hand; that's because there is no more F to accelerate it. I point this out to emphasize that in a free fall, only gravity and drag forces determine the net acceleration. The acceleration A of the toss to get up to U velocity has no bearing other than to get to U.
But if the object is just dropped, so that U = 0 initially, the initial acceleration will be a = g (1 - D/W) = g (1 - 0/W) = g because there is no initial drag force D ~ U^2 = 0. As the velocity builds up, so, too, will drag and the rate of build up, the acceleration will diminish.
Finally, if there is sufficient time of fall, the drag will build up to a point where it offsets the weight of the object. In other words, D = W and acceleration will be a = 0. When this happens, we say that V has reached terminal velocity as it will not get faster.
So there you are. Tossing the object will result in acceleration less than what it would have had it just been dropped. So at your measuring point, you should find the object is accelerating less with the tossed object. But as the tossed object starts with a head start on its velocity, it will pass your point of measuring faster than it would if just dropped.
- 4 years ago
In the aircraft reference frame, initial velocity is zero. Any measurement from aboard the aircraft will measure its initial speed to be zero, and assuming a stable level flight, that formula you mentioned H = -1/2*g*t^2 + v0*t + h, is what applies, as measured from the aircraft. In the ground reference frame, no, the initial velocity is not zero. The problem is a 2 dimensional projectile motion problem, and it begins with a HORIZONTAL speed equal to that of the aircraft. Horizontal position x is: x = v0x*t Vertical postion y is: y = -1/2*g*t^2 + h For the purpose of this problem, since they didn't tell you v0, you are solving it in the aircraft reference frame, or you are assuming a ground-stationary aircraft, like a helicopter hovering at a constant height and a constant position above the ground.