Question

A kangaroo jumps straight upward with a velocity of 5.77 m/s. Determine the height (to two decimal places) to which the kangaroo jumps.

Answer 1

Initial velocity of kangaroo in upward direction (u) = 5.77
`\((m)/(s)\)`

Final velocity at the highest point of the journey (v) = 0
`\((m)/(s)\)`

Let the maximum height reached be H.

Acceleration due to gravity = -9.8
`(m)/(s^2)\)`

Now, using the third equation of motion:

`\(v^2 = u^2 + 2aH\)`

`\(0^2 = 5.77^2 - 2 * 9.8 * H\)`

`\(5.77^2 = 2 * 9.8 * H\)`

`\(H = \((5.77^2)/(19.6) \)`

H = 1.698 m

The maximum height reached by kangaroo is 1.7 m.