Class work 1. A ball is kicked with an initial velocity if 40m/s from the ground at an angle of 30^0 to the horizontal. (Use g=10m/s^2 ) Calculate a) Horizontal and vertical component of initial velocity b) The vertical velocity after t=1s,2s,3s and 4s 2.1 KINEMATICS OF THE PARTICLE c) Position (overrightarrow (r)=xhat (i)+yhat (j)) after t=1s,2s,3s and 4s d) Time of flight (total time) e) maximum height f) Range of projectile 2. An air plane moving horizontally with velocity of 500km/hr at a height of 2km above the ground dropped a bomb when it directly above the target. By how much distance will the bomb miss the target? 3. An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 15.0m if her initial speed is 3.00m/s . What is the free -fall acceleration on the planet? 25

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Class work 1. A ball is kicked with an initial velocity if 40m/s from the ground at an angle of 30^0 to the horizontal. (Use g=10m/s^2 ) Calculate a) Horizontal and vertical component of initial velocity b) The vertical velocity after t=1s,2s,3s and 4s 2.1 KINEMATICS OF THE PARTICLE c) Position (overrightarrow (r)=xhat (i)+yhat (j)) after t=1s,2s,3s and 4s d) Time of flight (total time) e) maximum height f) Range of projectile 2. An air plane moving horizontally with velocity of 500km/hr at a height of 2km above the ground dropped a bomb when it directly above the target. By how much distance will the bomb miss the target? 3. An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 15.0m if her initial speed is 3.00m/s . What is the free -fall acceleration on the planet? 25

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1. a) The horizontal and vertical components of the initial velocity can be calculated using the equations: $V_x = V \cdot \cos(\theta)$ $V_y = V \cdot \sin(\theta)$ where $V$ is the initial velocity, $\theta$ is the angle of projection, and $V_x$ and $V_y$ are the horizontal and vertical components of the velocity, respectively. Given that $V = 40 \, \text{m/s}$ and $\theta = 30^\circ$, we can calculate: $V_x = 40 \cdot \cos(30^\circ) = 20 \, \text{m/s}$ $V_y = 40 \cdot \sin(30^\circ) = 20 \, \text{m/s}$ b) The vertical velocity after $t$ seconds can be calculated using the equation: $V_y = V_y + (-g) \cdot t$ where $V_y$ is the initial vertical velocity, $g$ is the acceleration due to gravity, and $t$ is the time. Given that $V_y = 20 \, \text{m/s}$, $g = 10 \, \text{m/s}^2$, and $t = 1s, 2s, 3s$, and $4s$, we can calculate: For $t = 1s$: $V_y = 20 + (-10) \cdot 1 = 10 \, \text{m/s}$ For $t = 2s$: $V_y = 20 + (-10) \cdot 2 = 0 \, \text{m/s}$ For $t = 3s$: $V_y = 20 + (-10) \cdot 3 = -10 \, \text{m/s}$ For $t = 4s$: $V_y = 20 + (-10) \cdot 4 = -20 \, \text{m/s}$ 2. c) The position of the projectile after $t$ seconds can be calculated using the equations: $x = V_x \cdot t$ $y = V_y \cdot t - \frac{1}{2} g t^2$ where $x$ and $y$ are the horizontal and vertical positions, respectively. Given that $V_x = 20 \, \text{m/s}$, $V_y = 20 \, \text{m/s}$, $g = 10 \, \text{m/s}^2$, and $t = 1s, 2s, 3s$, and $4s$, we can calculate: For $t = 1s$: $x = 20 \cdot 1 = 20 \, \text{m}$ $y = 20 \cdot 1 - \frac{1}{2} \cdot 10 \cdot 1^2 = 15 \, \text{m}$ For $t = 2s$: $x = 20 \cdot 2 = 40 \, \text{m}$ $y = 20 \cdot 2 - \frac{1}{2} \cdot 10 \cdot 2^2 = 20 \, \text{m}$ For $t = 3s$: $x = 20 \cdot 3 = 60 \, \text{m}$ $y = 20 \cdot 3 - \frac{1}{2} \cdot 10 \cdot 3^2 = 5 \, \text{m}$ For $t = 4s$: $x = 20 \cdot 4 = 80 \, \text{m}$ $y = 20 \cdot 4 - \frac{1}{2} \cdot 10 \cdot 4^2 = -20 \, \text{m}$ d) The time of flight can be calculated using the equation: $T = \frac{2V_y}{g}$ Given that $V_y = 20 \, \text{m/s}$ and $g = 10 \, \text{m/s}^2$, we can calculate: $T = \frac{2 \cdot 20}{10} = 4 \, \text{s}$ e) The maximum height can be calculated using the equation: $H = \frac{V_y^2}{2g}$ Given that $V_y = 20 \, \text{m/s}$ and $g = 10 \, \text{m/s}^2$, we can calculate: $H = \frac{20^2}{2 \cdot 10} = 20 \, \text{m}$ f) The range of the projectile can be calculated using the equation: $R

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