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

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)

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?

To determine how high the beanbag went, we can use the kinematic equation for vertical motion. The equation is:

h = (vi * t) - (1/2 * g * t^2)

Where:
h is the height
vi is the initial vertical velocity
t is the time
g is the acceleration due to gravity (approximately 9.8 m/s^2 on Earth)

In this case, the initial vertical velocity (vi) is 0 m/s since the beanbag was thrown vertically upwards. We can rewrite the equation as:

h = -(1/2 * g * t^2)

Now, we can substitute the given time (t = 3.3 seconds) into the equation:

h = -(1/2 * 9.8 * (3.3^2))
h = -(1/2 * 9.8 * 10.89)
h = -53.99

The negative sign indicates that the height is measured relative to the starting point (below the hand from where the beanbag was released). Therefore, the absolute value of the height would be 53.99 units. However, since we haven't specified the unit used for the time or height, we can't determine the exact height in a specific unit.