Question

You throw a beanbag in the air and catch it 3.3 seconds later. How high did it go?

Answer 1

time upward = 3.3/2 =1.65 s
v = Vi - 9.81 t
v = 0 at top
0 = Vi - 9.81 (1.65)
Vi = 16.2 m/s

h = Vi t - 4.9 t^2
h = (16.2)(1.65) - 4.9 (1.65)^2
h = 13.4 meters (good throw)

Answer 2

A 91.76-kg boater, initially at rest in a stationary 62.54-kg canoe, steps out of the canoe and onto the dock. If the boater moves out of the boat with a velocity of 3.52 m/s to the right, what is the final velocity of the boat?