Identify the initial vertical velocity component.
This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Here are the solutions to the questions.
Q8 (a): Show that the maximum height of a projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$.
Step 1: Identify the initial vertical velocity component.
The initial velocity $u$ is at an angle $\theta$ to the horizontal. The vertical component of the initial velocity is $u_y = u \sin \theta$.
Step 2: Apply a kinematic equation for vertical motion.
At the maximum height, the vertical velocity $v_y$ is $0$. Using the kinematic equation $v_y^2 = u_y^2 + 2a_y H$, where $a_y = -g$ (acceleration
