A squirrel tosses a acorn with a horizontal velocity of +2 m/s outward from a tree branch that is 8.0 m above

the ground.How long is the acorn falling through the air?What is the acorn's vertical component of velocity when it reaches the groung?How far from the point on the ground directly under the squirrel does the acorn land?A squirrel tosses a acorn with a horizontal velocity of +2 m/s outward from a tree branch that is 8.0 m above
To answer these questions, we must first try to understand why this squirrel is throwing away a perfectly fine acorn. Is the squirrel sick? Does he have rabies?





How much does the squirrel weigh, and how proficient is he at acorn tossing?





What is the radius of the tree branch? What type of tree is it?





What are the weather conditions?





Who is actually witnessing this amazing feat of squirrel dexterity?





What is the squirrel's name?





There are just too many questions surrounding this initial inquiry. I'm sorry, but next time have your material prepared before posting!!!








PS - Sticks and stones (and apparently acorns), hippo!A squirrel tosses a acorn with a horizontal velocity of +2 m/s outward from a tree branch that is 8.0 m above
The acorn never touches the land. Squirrels would never toss an acorn.
african or european?
using S = ut + 1/2 at.t vertically down


8.0 = 0 + 1/2 * 10* t.t (a= 10 m/s.s or 9.8m/s.s) [u = 0 because there is only a horizantal velocity the vertical velocity is 0 ]


8 = 5 * t.t


t.t = 1.6


t = 4/square root 10 (s) [the time taken to reach the ground]





using S = ut horizontally


S = 2* 4/ s.r. 10


S = 8/ s.r 10 (m) [the distance from the point of projection]





using V = u + at vertically down


V = 0 + 10*4/s.r.10


V = 40/ s.r.10 [ the vertical velocity when it reaches the ground]





the horizontal velocity is constant.





so the vertical component of velocity = _/1600/10 + 4 (using Pythagoras)


= _/160 + 4


= _/164





= 2_/41 (m/s)
thats a simple question. im bored. KE=PE
Vertical and horizontal components of motion are independent.


Vertical


s = ut + att/2


u = 0


So t = SQ RT 2S/a


t = SQ RT 2 x 8 /9.8 = 1.3s





v = u + at


u = 0


v = at = 9.8 x 1.3 = 12.7m/s





Horizontal


s = vt = 2 x 1.3 = 2.6m





The above ignores friction due to air resistance (drag)





PS Eskimo Joe is an idiot!