In a certain track and field event, the shotput has a mass of 7.30 kg and is released with a speed of 15.0 m> s at 40.0° above the horizontal over a competitor's straight left leg. what are the initial horizontal and vertical components of the momentum of this shotput

horizontal velocity = 11.5 m/s vertical velocity = 9.64 m/s. These are two elements to this problem. 1. Determining what to ignore. 2. Actually solving the problem. We really don't care what the shotput mass is, nor the relationship of the shot to the left leg of the competitor. The only things of interest are the velocity of the shotput and what angle above the horizontal it was thrown at. So, you simply need to multiply the total velocity by the sin and cosine of the angle to get the respective vertical and horizontal velocities. So: Vvert = sin(40.0°) * 15.0 = 0.64278761 * 15 = 9.641814145 Vhorz = cos(40.0°) * 15.0 = 0.766044443 * 15 = 11.49066665 Rounding the respective velocities to 3 figures gives a horizontal velocity of 11.5 m/s and a vertical velocity of 9.64 m/s.

Wind or electric energy force force of friction