Question

1. cuban tree frogs are skilled jumpers. they can launch themselves at 4.5 meters per second at an angle of 26° above horizontal. assuming the frog is jumping on level ground, how far will the frog travel in a single jump? 0.95 m 1.6 m 3.3 m 2.0 m

Explanation:

Step1: Find the time of flight

The vertical component of the initial velocity is \(v_{0y}=v_0\sin\theta\), where \(v_0 = 4.5\ m/s\) and \(\theta=26^{\circ}\). Using the equation \(y = v_{0y}t-\frac{1}{2}gt^{2}\), since \(y = 0\) (level - ground), \(0=v_0\sin\theta t-\frac{1}{2}gt^{2}\). Factoring out \(t\) gives \(t = 0\) (initial time) or \(t=\frac{2v_0\sin\theta}{g}\). Substituting \(g = 9.8\ m/s^{2}\), \(v_0 = 4.5\ m/s\), and \(\theta = 26^{\circ}\), we have \(t=\frac{2\times4.5\times\sin(26^{\circ})}{9.8}\).
\(\sin(26^{\circ})\approx0.4384\), so \(t=\frac{2\times4.5\times0.4384}{9.8}=\frac{4.5\times0.8768}{9.8}\approx0.4\ s\)

Step2: Find the horizontal distance

The horizontal component of the initial velocity is \(v_{0x}=v_0\cos\theta\). The horizontal distance \(x = v_{0x}t\), and \(v_{0x}=v_0\cos\theta\) with \(v_0 = 4.5\ m/s\) and \(\theta = 26^{\circ}\), \(\cos(26^{\circ})\approx0.8988\). So \(x=v_0\cos\theta\times t\). Substituting \(v_0 = 4.5\ m/s\), \(\cos(26^{\circ})\approx0.8988\), and \(t\approx0.4\ s\), we get \(x = 4.5\times0.8988\times0.4\approx1.6\ m\)

Another way: Use the range formula \(R=\frac{v_0^{2}\sin(2\theta)}{g}\)
Since \(v_0 = 4.5\ m/s\), \(\theta = 26^{\circ}\), then \(2\theta=52^{\circ}\), \(\sin(52^{\circ})\approx0.7880\)
\(R=\frac{4.5^{2}\times0.7880}{9.8}=\frac{20.25\times0.7880}{9.8}=\frac{16.057}{9.8}\approx1.6\ m\)

Answer:

1.6 m