Question

A softball is thrown from the origin of an x-y coordinate system with an initial speed of 18 m/s at an angle of 35∘ above the horizontal. Part A Find the x positions of the softball at the times t = 0.50 s, 1.0 s, 1.5 s, and 2.0 s.

Answer 1

What we need to first do is split the ball's velocity into vertical and horizontal components. To do that multiply by the sin or cos depending upon if you're looking for the horizontal or vertical component. If you're uncertain as to which is which, look at the angle in relationship to 45 degrees. If the angle is less than 45 degrees, the larger value will be the horizontal speed, if the angle is greater than 45 degrees, the larger value will be the vertical speed. So let's calculate the velocities sin(35)*18 m/s = 0.573576436 * 18 m/s = 10.32437585 m/s cos(35)*18 m/s = 0.819152044 * 18 m/s = 14.7447368 m/s Since our angle is less than 45 degrees, the higher velocity is our horizontal velocity which is 14.7447368 m/s. To get the x positions for each moment in time, simply multiply the time by the horizontal speed. So 0.50 s * 14.7447368 m/s = 7.372368399 m 1.00 s * 14.7447368 m/s = 14.7447368 m 1.50 s * 14.7447368 m/s = 22.1171052 m 2.00 s * 14.7447368 m/s = 29.48947359 m Rounding the results to 1 decimal place gives 0.50 s = 7.4 m 1.00 s = 14.7 m 1.50 s = 22.1 m 2.00 s = 29.5 m