Fundamentals of Physics,Vol 1 (Chapters 1 - 20)

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
43. We neglect air resistance for the duration of the motion (between “launching” and “landing”), so a = g = 9 . 8m/s 2 (we take downward to be the y direction). We use the equations in Table 2-1 (with y replacing ∆ x ) because this is a = constant motion. (a) At the highest point the velocity of the ball vanishes. Taking y 0 =0,weset v =0in v 2 = v 2 0 2 gy , and solve for the initial velocity: v 0 = 2 .S in c e y = 50 m we Fnd v 0 =31m/s. (b) It will be in the air from the time it leaves the ground until the time it returns to the ground ( y = 0). Applying Eq. 2-15 to the entire motion (the rise and the fall, of total time t> 0) we have y = v 0 t 1 2 gt 2 = t = 2 v 0 g which (using our result from part (a)) produces t =6 . 4 s. It is possible to obtain this without using part (a)’s result; one can Fnd the time just for the rise (from ground to highest point) from
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online