2022(9.8 m/s)(0.76 m)3.86 m/s .vgy===This is a consequence of Eq. 2-16 where velocity vvanishes. As the player reaches y1= 0.76 m – 0.15 m = 0.61 m, his speed v1satisfies vvgy021212−=, which yields 2221012(3.86 m/s)2(9.80 m/s)(0.61 m)1.71 m/s .vvgy=−=−=The time t1that the player spends ascendingin the top Δy1= 0.15 m of the jump can now be found from Eq. 2-17: ()()111120.15 m10.175 s21.71m/s0yvv ttΔ=+¡==+which means that the total time spent in that top 15 cm (both ascending and descending) is 2(0.175 s) = 0.35 s = 350 ms. (b) The time t2when the player reaches a height of 0.15 m is found from Eq. 2-15:
This is the end of the preview.
access the rest of the document.
Velocity, Quadratic equation, 2006 albums, Reach, Ascending and Descending