Question

14 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, what is the maximum height above the ground that the frog will achieve? 0.61 m 0.05 m 0.20 m 1.2 m

Explanation:

Step1: Find the vertical component of the initial velocity

The initial velocity \(v_0 = 4.5\ m/s\) and the launch angle \(\theta=26^{\circ}\). The vertical component of the initial velocity is \(v_{0y}=v_0\sin\theta\).

$$v_{0y}=4.5\times\sin(26^{\circ})$$

$$v_{0y}=4.5\times0.4384\approx1.973\ m/s$$

Step2: Use the kinematic equation for vertical motion

The kinematic equation \(v_y^{2}=v_{0y}^{2}-2gh\) (at maximum height \(v_y = 0\)). Solving for \(h\) (maximum height), we get \(h=\frac{v_{0y}^{2}}{2g}\) (where \(g = 9.8\ m/s^{2}\)).

$$h=\frac{(1.973)^{2}}{2\times9.8}$$

$$h=\frac{3.893}{19.6}\approx0.20\ m$$

Answer:

\(0.20\ m\) (the third option)