Question

Dr. Bennett releases a bowling ball with an initial velocity of 2.9 m/s; the ball slows down with a constant acceleration of -0.28 m/s.

Answer 1

Vo=2.9 m/s a=-0.28 m/s AX= 2 v=0 v²=v8 tzada o = 2.99 + 2(-:28) Ax Da=-(209²) -0.48 sx= 17.2m

by V = Vot at t = V- Vo -2.9 a -o-28 il 10-36 seconds

cs Vo=2.9 m/s DX =17:2 g.bm a=-0.28 m/s² V=? 8.3) va = v. + 2 asal (2-4)? + (2x -0.28 x 3.59 V = J 3.59 -1.89 m/s =

d , V = votat 1.89 = 2.9 +(-0.28) + t = 3.65econds