# Projectile Motion Problem Solution Let's break down the problem step by step: --- ## Problem Statement - **Projectile is shot from a 135 m high cliff** - **Initial speed:** 60 m/s - **Angle:** 30° above horizontal - **Gravity \( g \):** 10 m/s² - **Given:** \( \sin 30^\circ = 0.5 \), \( \cos 30^\circ = 0.87 \) --- ## (a) Initial Velocity in Vector Component Form Initial velocity \( \vec{v}_i \): \[ v_{ix} = v_i \cos \theta = 60 \times 0.87 = 52.2 \text{ m/s} \] \[ v_{iy} = v_i \sin \theta = 60 \times 0.5 = 30 \text{ m/s} \] **Vector form:** \[ \boxed{\vec{v}_i = (52.2\, \hat{i} + 30\, \hat{j}) \text{ m/s}} \] --- ## (b) Time to Hit the Ground Use the kinematic equation for vertical motion: \[ \Delta y = v_{iy} t + \frac{1}{2} a t^2 \] Here, - \( \Delta y = -135 \) m (downwards) - \( v_{iy} = 30 \) m/s (upwards) - \( a = -10 \) m/s² (downwards) Plug in the values: \[ -135 = 30t - 5t^2 \] \[ 5t^2 - 30t - 135 = 0 \] \[ t^2 - 6t - 27 = 0 \] Solve the quadratic equation: \[ t = \frac{6 \pm \sqrt{36 + 108}}{2} = \frac{6 \pm \sqrt{144}}{2} = \frac{6 \pm 12}{2} \] Choose the positive root: \[ t = \frac{6 + 12}{2} = \frac{18}{2} = 9 \text{ s} \] **Time to hit the ground:** \[ \boxed{9 \text{ s}} \] --- ## (c) Distance \( X \) from Base of Cliff Horizontal motion: \[ X = v_{ix} t = 52.2 \times 9 = 469.8 \text{ m} \] **Distance from base:** \[ \boxed{469.8 \text{ m}} \] --- ## **Summary of Answers** **(a)** \( \vec{v}_i = (52.2\, \hat{i} + 30\, \hat{j}) \) m/s **(b)** \( t = 9 \) s **(c)** \( X = 469.8 \) m