6/25/2023 0 Comments Projectile motion lab![]() And since the starting and ending points have the same elevation, we can then assume that the projectile has equal speed at those two points. We assume this to be true since we are also assuming that there is no air resistance. So we choose the final velocity to be just before it hits the ground.Īnd what is the final velocity before it hits the ground? Well, the projectile does not lose any energy while from the time right after it is launched to the time just before it lands. Fortunately, this problem can be solved just with the motion of the projectile before it hits the ground, so we don't need to concern ourselves with anything after that. Then only after it hits the ground will it have zero velocity, but hitting the ground will introduce another force to this system, and we would need to use more equations to describe its motion. Just before it hits the ground, the projectile has some downward speed. So we should only apply them to the motion of the projectile right after it is thrown and right before it hits the ground. This means that the only force acting on it is the force of gravity. Gray footnotes are citations and links to outside references.The equations that we are using to solve this problem only apply when the projectile is in free fall. Hovering over these bubbles will make a footnote pop up. Taking the ball as a sphere of radius ~1cm and mass ~10g moving through the air (density ~1kg/m^3) at ~1m/s, how big is the force of air resistance? How does this compare to the force of gravity? What does this tell you about the relevance of air resistance as a systemic error? ![]() Instead, we will measure the effective diameter, \(d_\text\) is the density of air, \(v\) is the velocity of the object (relative to the air), and \(A\) is the cross-sectional area of the object. However, we cannot rely on that assumption. If the photogate were exactly halfway down the ball, then this distance would be the diameter of the ball. Velocity is change in position over change in time, so we need to determine first how far the ball will move when it crosses the photogate. We are going to determine the velocity by measuring the time \(\delta t\) it takes the ball to cross in front of the photogate. Part II: Determining the Effective Diameter of the Steel Ball If so, press "Stop Collection" (same button) 1 - everything is set up correctly, and you can move on to the next part. Roll the ball down the hill and ensure that it shows two times: one at which the ball entered the photogate, and one at which the ball left it. On the computer, open "Exp3_t1_t2" click "Connect" on the box that pops up.Ĭlick the green "Start Collection" button at the top of the screen to start "recording" the output of the photogate. ![]() Plug it into "DIG/SONIC 1" port of the "LabPro" on the wall. Mount it onto the stand at the side of the ramp, and ensure that the beam crosses that track at approximately the center of the ball when the ball is placed at the relevant position. Estimate an uncertainty in this distance \(h\) as well. Measure the distance \(h\) vertically from the bottom of the ball to the floor, using your plumb bob as a guide to a perfect vertical. ![]() Place the ball right at the end of the ramp (you may have to hold it there). This setup of having a hanging mass to show you a vertical line is called a plumb bob. This gives you a purely-vertical line from the bottom of the ramp to the floor directly below (which will allow us to measure horizontal distance precisely). Ensure that it is not resting on the floor, but instead hanging a short distance (a cm or two) above it. Hang the paper clip from the end of the ramp. Clamp it down so that the position is fixed. Place your ramp so that the front edge of the platform lines up with the edge of the table (so that when the ball rolls off the ramp, it goes straight off the edge).
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