You are floating down a river at a speed of 3.0 m/s in a canoe while drinking a

Question

You are floating down a river at a speed of 3.0 m/s in a canoe while drinking a can of soda. You
see your friend on a bridge ahead whose arm is 15 m above the water. You throw a can of soda
straight up from the canoe relative to yourself (remember, the canoe and you are in motion). The
soda can rises and, at the exact top of its trajectory, is caught by your friend on the bridge.
4 points) How long was the can in the air?
(4 points) How far from the bottom of the bridge was the canoe when the can was launched?

Explanation / Answer

a) top of trajectory -> vertical velocity is zero; the horizontal velocity does not change vh = 3m/s
b) initial horizontal speed = 3.0m/s, to get vertical you need the distance the can rose from canoe to catch. If you assume it was thrown from water level near the bottom of the canoe then the distance is 15m and you can get the initial vertical speed by conservation of energy: 1/2 m v^2 = mgh
v = sqrt(2gh)
c) time in the air = time to reduce the vertical velocity to zero: gt = v, t = v/g where is the vertical velocity just found
d) you are not given enough initial information to decide this. in part b I had to assume a starting position for the can to get a solution. Here there are too many things left to guess like what was the distance between the friend's hand and the bridge and what is the height of the canoe above the water.

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