Question

question if the initial angle were smaller, with all the other quantities remaining the same, the kinetic energy of the grasshopper at its highest point would be: greater than before less than before unchanged if the mass of the grasshopper is m, and its maximum height h, with θ the initial angle, the change in kinetic energy of the grasshopper between the start of its jump and its greatest height has magnitude: zero mgh cosθ mgh independent of initial angle how then would a smaller initial angle change the initial kinetic energy, and the initial speed, that the grasshopper needs to reach the same height of 1 meter? the grasshoppers jump would be at the same speed. the grasshopper must jump faster, and therefore with more kinetic energy. the grasshopper can jump more slowly, and therefore with less kinetic energy. practice it use the worked example above to help you solve this problem. a powerful grasshopper launches itself at an angle of 45° above the horizontal and rises to a maximum height of 1.09 m during the leap. (see the figure.) with what speed v₁ did it leave the ground? neglect air resistance. m/s exercise hints: getting started | im stuck! a catapult launches a rock at a 48° angle with respect to the horizontal. find the maximum height attained if the speed of the rock at its highest point is 25.2 m/s. m

Explanation:

Step1: Analyze vertical - motion at highest point

At the highest point of the grass - hopper's jump, the vertical component of velocity \(v_y = 0\). The initial velocity components are \(v_{0x}=v_0\cos\theta\) and \(v_{0y}=v_0\sin\theta\), where \(v_0\) is the initial speed and \(\theta\) is the initial angle. The kinetic energy at the highest point is \(K=\frac{1}{2}mv_{0x}^2=\frac{1}{2}m(v_0\cos\theta)^2\). If \(\theta\) decreases (while \(v_0\) remains the same), \(\cos\theta\) increases, so the kinetic energy at the highest point is greater than before.

Step2: Use conservation of mechanical energy

The change in mechanical energy \(\Delta E=\Delta K+\Delta U = 0\) (since there is no non - conservative force, neglecting air resistance). The initial kinetic energy is \(K_0=\frac{1}{2}mv_0^2\) and at the highest point \(K=\frac{1}{2}mv_{0x}^2\) and \(U = mgh\). So, \(\Delta K=K - K_0=-mgh\), and the magnitude of the change in kinetic energy between the start and the greatest height is \(mgh\), independent of the initial angle.

Step3: Analyze reaching the same height with a different angle

To reach a height \(h\), we use \(v_{0y}^2 = 2gh\) (from \(v_y^2 - v_{0y}^2=- 2gh\) with \(v_y = 0\) at the highest point). \(v_{0y}=v_0\sin\theta\). If \(\theta\) is smaller, to have the same \(v_{0y}\) (to reach the same height \(h\)), \(v_0\) must be larger. So the grass - hopper must jump faster and with more kinetic energy.

Step4: Solve for the initial speed of the grass - hopper

Using \(v_{0y}^2 = 2gh\), where \(v_{0y}=v_0\sin\theta\). Given \(\theta = 45^{\circ}\) and \(h = 1.09m\), and \(v_{0y}^2 = 2gh\), so \(v_0\sin\theta=\sqrt{2gh}\). Then \(v_0=\frac{\sqrt{2gh}}{\sin\theta}\). Substituting \(g = 9.8m/s^2\), \(h = 1.09m\) and \(\theta = 45^{\circ}\), we have \(v_0=\frac{\sqrt{2\times9.8\times1.09}}{\sin45^{\circ}}=\frac{\sqrt{21.364}}{\frac{\sqrt{2}}{2}}\approx\frac{4.62}{0.707}\approx6.53m/s\).

Step5: Solve for the maximum height of the rock

At the highest point, the horizontal component of velocity \(v_x = v_0\cos\theta\). Given \(v_x = 25.2m/s\) and \(\theta = 48^{\circ}\), we can find \(v_0=\frac{v_x}{\cos\theta}\). Then, using \(v_{0y}=v_0\sin\theta\) and \(v_{0y}^2 = 2gh\), we first find \(v_0=\frac{25.2}{\cos48^{\circ}}\approx\frac{25.2}{0.669}\approx37.67m/s\). Then \(v_{0y}=v_0\sin\theta=37.67\times\sin48^{\circ}\approx37.67\times0.743\approx28m/s\). Using \(v_{0y}^2 = 2gh\), we get \(h=\frac{v_{0y}^2}{2g}=\frac{28^2}{2\times9.8}=\frac{784}{19.6}=40m\).

Answer:

1. greater than before
2. \(mgh\) independent of initial angle
3. The grass - hopper must jump faster, and therefore with more kinetic energy.
4. \(6.53\)
5. \(40\)