A kangaroo can jump straight up about 2.1 m. What is its takeoff speed?

Question

_____ m/s

Answer 1

u need more info like weight and the work used

Answer 2

The kinetic energy associated with the kangaroo's leap gets converted to potential energy.

KE at takeoff = (1/2) * m * v^2

PE at maximum height = m * g * h

m * g * h = (1/2) * m * v^2

The m's cancel, leaving

v^2 = g*h / (1/2)

or

v = squareroot (2 * g * h)

h = 2.1 m; g = 9.8 m/s^2

so

v = squareroot (2 * 9.8 m/s^2 * 2.1m)
= 6.4 m/s

Answer 3

All take off speed starts with Zero... well after that it may change.