A squirrel sees a BIG acorn being thrown from a tree by a chipmunk and attempts to snatch it. The acorn is thrown from a 10.0 m high branch and with a horizontal velocity of 1.1 m/s. The squirrel on the ground is 5.0 m from the base of the tree. What is the of the squirrel’s average velocity if it grabs the acorn in its mouth just as it touches the ground?


Answer 1
Answer: Since the acorn is thrown horizontally, the initial vertical velocity is 0 m/s. Let’s use the following equation to determine the time to fall 10 meters. 

d = vi * t + ½ * a * t^2, vi = 0, a = 9.8 
10 = 4.9 * t^2 
t = √(10/4.9) 
This is approximately 1.43 seconds. Let’s use the following equation to determine the horizontal distance it moves. 

d = v * t = 1.1 * √(10/4.9) 
This is approximately 1.57 meters. To determine the horizontal distance between the acorn and squirrel, subtract this distance from 5 meters. 

d = 5 – 1.1 * √(10/4.9) 

This is approximately 3.83 meters. To catch the acorn, the squirrel must move this distance in the same time as the acorn falls 10 meters. 

Average velocity = ([5 – 1.1 * √(10/4.9] ÷ √(10/4.9) = 2.4 m/s 

I hope this helps you to understand how to solve this type of problem.
Answer 2


For big answers like this I write it down on a piece of paper but I don't know how to put it into my answer :/

The average velocity of the squirrel is 2.4 m/s.

I hope it helps!

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