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 $\(\frac{m}{s}\)$

Final velocity at the highest point of the journey (v) = 0 $\(\frac{m}{s}\)$

Let the maximum height reached be H.

Acceleration due to gravity = -9.8$\frac{m}{s^2}\)$

Now, using the third equation of motion:

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

$\(0^2 = 5.77^2 - 2 \times 9.8 \times H\)$

$\(5.77^2 = 2 \times 9.8 \times H\)$

$\(H = \(\frac{5.77^2}{19.6} \)$

H = 1.698 m

The maximum height reached by kangaroo is 1.7 m.