Identify the initial vertical velocity component.

April 10, 2026Physics

ScanSolve AI Answer

April 10, 2026

Answer

\frac{u^2 \sin^2 \theta}{2g}$.

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

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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 = (u^2 ^2 )/(2g).
The initial velocity u is at an angle to the horizontal.
The vertical component of the initial velocity is u_y = u.

Asked on April 10, 2026Physics

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.

ScanSolve AI AnswerAnswered on April 10, 2026

Here are the solutions to the questions. Q8 (a): Show that the maximum height of a projectile is given by H = (u^2 ^2 )/(2g). Step 1: Identify the initial vertical velocity component. The initial velocity u is at an angle to the horizontal. The vertical component of the initial velocity is u_y = u . 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