**99 points** i will give brainliest answer! explain answer! 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?

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.