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, how far will the kangaroo travel in a single bounce? 11.9 m 6.12 m 8.15 m 7.36 m

Explanation:

Step1: Find the time of flight

The vertical component of the initial velocity is \(v_{0y}=v_0\sin\theta\), where \(v_0 = 8.50\ m/s\) and \(\theta = 45^{\circ}\). Using the kinematic equation \(y = v_{0y}t-\frac{1}{2}gt^{2}\), when the kangaroo lands \(y = 0\). So \(0=v_0\sin\theta t-\frac{1}{2}gt^{2}\), \(t=\frac{2v_0\sin\theta}{g}\) (since \(t = 0\) is the initial time). Substituting \(v_0 = 8.50\ m/s\), \(\theta = 45^{\circ}\), and \(g = 9.8\ m/s^{2}\), \(t=\frac{2\times8.50\times\sin45^{\circ}}{9.8}\).

Step2: Find the horizontal distance

The horizontal component of the velocity is \(v_{0x}=v_0\cos\theta\). The horizontal distance \(x = v_{0x}t\). Since \(v_{0x}=v_0\cos\theta\) and \(t=\frac{2v_0\sin\theta}{g}\), then \(x=\frac{v_{0}^{2}\sin2\theta}{g}\) (using the double - angle formula \(\sin2\theta=2\sin\theta\cos\theta\)). Substituting \(v_0 = 8.50\ m/s\), \(\theta = 45^{\circ}\) (so \(\sin2\theta=\sin90^{\circ}=1\)) and \(g = 9.8\ m/s^{2}\), \(x=\frac{8.50^{2}\times1}{9.8}=\frac{72.25}{9.8}\approx7.37\ m\)

Answer:

7.36 m (due to rounding differences in intermediate steps, the closest value to the calculated \(x\approx7.37\ m\) is 7.36 m)