Max KE of a squirrel jumping from the top of a tree

[Boop de Boop]

Homework Statement

1. A squirrel jumps horizontally from the top of a 25 m tall tree with a speed of 8 m/s. What is the maximum kinetic energy of the squirrel during it`s entire trip down to the ground from the top of the tree?
2.If another squirrel of the same mass jumps straight upward from the ground well enough to get 4m off the ground and then falls back to the ground, what is the kinetic energy in the following locations and time:
a)3m above ground, on the way up
b)2m above ground, on the way down
c)1m above ground, on the way down

Homework Equations

Pe:mgh
ke:0.5mv^2
pei+kei=pef+kef

The Attempt at a Solution

I know that we have to use the ke initial and add it with the new kinetic energy that is gained during gthe trip in order to calculate total kinetic energy before it hits the ground for prolem 1. But I am really confused and would appreciate guidance. I do not need the answers just help. :)

 

[haruspex]

I know that we have to use the ke initial and add it with the new kinetic energy that is gained during gthe trip in order to calculate total kinetic energy before it hits the ground for prolem 1

So where are you stuck, finding the initial KE, finding the KE gained in the fall, or in adding the two?
Can't tell what your confusion is without seeing your attempt.

 

[Boop de Boop]

I used the kinematic equation vf^2=vi^2+2ad and found final velocity which is 23.74. Then I plugged into ke-)0.5mv^2. I'm confused with kinetic initial. Is it just ke formula but v final is 8

 

[haruspex]

I used the kinematic equation vf^2=vi^2+2ad and found final velocity which is 23.74. Then I plugged into ke-)0.5mv^2. I'm confused with kinetic initial. Is it just ke formula but v final is 8

That kinetic equation is usually used for one dimensional motion, but here the initial velocity and the acceleration are in different directions.
However, it will give the right result here thanks to Pythagoras; see the later Edit below. Suppose we take that equation and multiply it everywhere by ½m:
½mv_f²=½mv_i²+mad, where d is the height of the descent and a is g. You can see that this represents
Final KE = initial KE + lost GPE.
I.e. it is the equation for conservation of mechanical work.

Edit: if you want to do it by kinetics rather than work conservation, a more usual path would be to treat the horizontal and vertical separately:
v_fx²=v_ix²+0 (no horizontal acceleration)
v_fy²=v_iy²+2(-d)(-g), where v_iy=0.
Then you can add these to get the square of the landing speed.

 

[Apashanka]

A squirrel jumps horizontally from the top of a 25 m tall tree with a speed of 8 m/s. What is the maximum kinetic energy of the squirrel during it`s entire trip down to the ground from the top of the tree?

From energy conservation KE_height+mgh=KE_ground+0 where KE_ground is the max. KE

 

[Apashanka]

2.If another squirrel of the same mass jumps straight upward from the ground well enough to get 4m off the ground and then falls back to the ground, what is the kinetic energy in the following locations and time:

From energy conservation
KE_i+0=KE_height+mgh ,
During upward journey.
During downward journey
mgh_max=mgh+KE_height

 

[haruspex]

From energy conservation KE_height+mgh=KE_ground+0 where KE_ground is the max. KE

@Boop de Boop seems to understand that (see post #1), but became confused when trying to solve it with SUVAT equations (see post #3).
By the way @Boop de Boop, you meant "kinetic", not "kinematic". Kinematics concerns the geometry of moving systems. See e.g. https://en.m.wikipedia.org/wiki/Kinematics.