Calculating the Force of a Kangaroo's Legs

[Shoebox]

Kangaroos have large, powerful legs for jumping. A male kangaroo that has a mass of 66.5 kg can accelerate to a vertical velocity of 6.08 before his feet leave the ground (at a height of 1m). What force is the kangaroo's legs able to exert on the ground in order to do this?

 

[TESL@]

Assuming that the height is 0 initially,

6.08^2 = 2a
F = 66,5 (a + 9,81)

 

[Suraj M]

You have the final velocity and also the distance covered before attaining that velocity, so use 3rd equation of motion and find acceleration, i think you can do the rest!

 

[Shoebox]

thank you! my only question is, how did you know this part: 6.08^2=2a?

 

[Suraj M]

v²-u²=2as and your u=0 and s=1

 

[TESL@]

Timeless velocity equation: v(final)^2 = 2ax if there is uniform acceleration. It is easy to derive from a simple velocity vs. time graph.

 

[Shoebox]

and now to figure out how high he can jump. I am not sure what to use, because f=ma does not deal with distance

 

[TESL@]

Use the same formula.

 

[Shoebox]

the formula for the previous is the same as vf^2=vi^2+2ax, correct?

 

[Suraj M]

f=ma

You know both the quantities on the RHS.

 

[Suraj M]

the formula for the previous is the same as vf^2=vi^2+2ax, correct?

yes

 

[Shoebox]

okay.. 6.08^2=2(18.48)(x)
im getting x=1..where am i going wrong?

 

[Suraj M]

No, you found $a$ from the equation v²=2ax... ;you already know$x=1 m$ you should find $a$ , which you have done very nicely,now substitute in the formula $F=ma$

 

[Shoebox]

i found the force to be 1881. Now i am trying to find how far he can jump, or x

 

[Suraj M]

For that you need to use a=g and v_intial= 6.08 units, because the kangaroo is in the air and gravity is acting on it.
substitute in the equation$v²_{final}-v²_{initial}=2ax$ be careful while you choose the sign for $a$ when you substitute.

 

[Shoebox]

so my equation setup would be:
(0)^2 - (6.08)^2=2(18.48)(x)

 

[Suraj M]

how far he can jump, or x

how far, or how high??

 

[Shoebox]

high

 

[Suraj M]

so my equation setup would be:
(0)^2 - (6.08)^2=2(18.48)(x)

choose your$a$ well, its in the air, gravity acts on it so it should be $a=g=-9.8 m/s²$ can you tell me why the -ve sign?

 

[Shoebox]

thats right, gravity acts on him once he jumps. okay i solved (0)^2 - (6.08)^2=2(9.8)(x) and got x=1.886

 

[Shoebox]

the -9.8 is negative because of the direction of his acceleration.

 

[Suraj M]

Good, but i don't see you use it here

. okay i solved (0)^2 - (6.08)^2=2(9.8)(x) and got x=1.886

your answer's right!