G a clear explanation of how to get them and drawing theta = 10.0 deg vi = 7.00
ID: 2133150 • Letter: G
Question
G a clear explanation of how to get them and drawing
theta = 10.0 deg vi = 7.00 mls Suppose that during the 1-D Motion lab, you reset the ramp angle, removed the motion detector, and launched the cart from the bottom end with an initial speed Vi, just as in lab, but this time you launched it much faster so it left the end of the track and was projected into the air and crashed onto the floor. For this problem, it is reasonable to ignore the effects of air resistance. We can also ignore any friction with the track. The cart accelerated through a distance of 0. 850 m along the track before being projected into the air. The vertical distance from the end of the track to the floor is 1.00 m as shown in the diagram. The remaining numerical values used in this problem will be unique to each student. Determine the acceleration of the cart along the surface of the ramp. Determine the speed of the cart at the instant that it leaves the track. Draw the velocity vector for the cart at the instant it leaves the track using a non-rotated coordinate system (that is, the x-axis is parallel to the floor and y axis is vertical). Determine the horizontal and vertical components of the velocity vector at the instant the cart leaves the track. Express the velocity vector from Part C using unit vector notation. Determine how long the cart is in free-fall. That is, determine the time interval from the instant that the cart loses contact with the track to the instant that it makes contact with the floor. Determine the horizontal displacement of the cart, as measured from the end of the track (along the floor). Draw the velocity vector for the cart at the instant it makes contact with the floor. Calculate the velocity components, and the speed and direction (with respect to the +x-axis).Explanation / Answer
final speed:
vy = vy0 - gt = 1.179 - 9.8 * 0.58781 =-4.581538 = -4.58 m/s
vx = 6.687 m/s
speed = sqrt(6.687*6.687+4.581538*4.581538) = 8.11 m/s
theta = 360 - arctan(4.581538/6.687) = 326 degrees
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