Question

1. when the red kangaroo of australia is moving at full speed, it can bounce at a speed of 8.50 meters per second, at an angle of 45.0° above horizontal. assuming the kangaroo is bouncing on level ground, what is the maximum height above the ground that the kangaroo will achieve?

1.84 m
8.15 m
2.10 m
7.36 m

Explanation:

Step1: Find the vertical component of the initial velocity

The initial velocity \(v = 8.50\ m/s\) and the angle \(\theta=45.0^{\circ}\). The vertical component of the initial velocity is \(v_{0y}=v\sin\theta\).

$$v_{0y}=8.50\times\sin(45^{\circ})=8.50\times\frac{\sqrt{2}}{2}\approx6.01\ m/s$$

Step2: Use the kinematic equation for vertical motion

The kinematic equation for vertical motion \(v_{y}^{2}=v_{0y}^{2}- 2gh\). At the maximum - height, \(v_{y} = 0\). We want to solve for \(h\).

$$h=\frac{v_{0y}^{2}}{2g}$$

Substitute \(v_{0y}\approx6.01\ m/s\) and \(g = 9.8\ m/s^{2}\) into the formula.

$$h=\frac{(6.01)^{2}}{2\times9.8}=\frac{36.1201}{19.6}\approx1.84\ m$$

Answer:

1.84 m