Recent content by Shoebox

i am not.. but using a kinematics chart works.

it is 1.88 plus one. I just forgot to add the 1m distance he had already covered

i thought the equation would give me his change in distance. i do not see where i can fit the force into the problem, which is why i am asking for help

it reads pounds. so it would be 183.26 pounds. but i still don't understand how the acceleration would affect his weight

so i could use the equation for the x direction: a=f/m. for 3-5 seconds that gave me an acceleration of -0.366 m/s2. now I am stuck

I am trying to find the highest distance Mr. Kangaroo can jump. But apparently my answer of 1.88 m is incorrect, and i do not understand where i went wrong.

i was trying to find the velocity by multiplying the mass by force over time. looks like i got my variables mixed up. but i realize that this is not correct. how about using F/m=a... the force at 6 seconds is 0. the mass is 5.46.. but 0/5.46 gives me 0 acceleration

it would read 821 N

Homework Statement man's weight is 83.8 kg 1lb=4.48N what would the scale read (in N) if he went from 0m/s to 7m/s up in 6.56 seconds? Homework Equations F=ma a=v/tThe Attempt at a Solution a=7/6.56 a=1.07 f=(83.5)(1.07) f= 89.3N

Homework Statement 5.46 kg object moves along the x-axis. It is starting from rest. What is the velocity at 6 seconds? Homework Equations F=ma a=f/m The Attempt at a Solution x=5.46(v/6)[/B]

Homework Statement How high can a kangaroo jump. He is 66.5 kg, exerts 1881 N force on the ground when jumping, and reaches a velocity of 6.08 before leaving the ground (at 1 m). a=18.48 Homework Equations F=ma The Attempt at a Solution looking for distance. vf^2=vi^2+2ax 36.97=2(9.8)x x=1.88

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

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

high

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