Activity 3 Projectile Motion 1.) In a mission to destroy a target an F- 18A level at an altitude of 1000 m with a velocity 600 km/h uses the glide bomb to attack terrorist group's chemical weapon factory what horizontal distance in km cr Hornet release the bomb so a range should the as to hit the target. 2.) If a car that was pushed over the cliff had an intial velocity of 5.0 m/s in the horizontal direction, ? Ca) How long would the car be in the air cb) How far would the car land away from the cliff (c) what is the car's final velocity?
Answer (1)
Answer:(a) Time in the air: 2.02 seconds(b) Horizontal distance: 10.1 meters(c) Final velocity: 20.4 m/sQuestion 1: F-18A Bomb DropTo solve for the horizontal distance, we use the following information:Altitude [tex]h = 1000 \, \text{m}[/tex]Horizontal velocity [tex]v_x = 600 \, \text{km/h} = 166.67 \, \text{m/s}[/tex]Acceleration due to gravity [tex]g = 9.8 \, \text{m/s}^2[/tex]1. Find the time of flight [tex](\( t \))[/tex]: The bomb is in free fall, so we calculate the time to hit the ground: [tex]h = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2h}{g}}[/tex] Substituting the values:[tex]t = \sqrt{\frac{2(1000)}{9.8}} \approx 14.28 \, \text{seconds}[/tex]2. Find the horizontal distance [tex](\( d \))[/tex]: Since horizontal motion is uniform:[tex]d = v_x \cdot t[/tex] Substituting the values:[tex]d = 166.67 \cdot 14.28 \approx 2380 \, \text{meters} = 2.38 \, \text{km}[/tex]Answer: The F-18A should release the bomb at a horizontal distance of approximately 2.38 km from the target.Question 2: Car Over the CliffWe know:Initial horizontal velocity [tex]v_x = 5.0 \, \text{m/s}[/tex]Acceleration due to gravity [tex]g = 9.8 \, \text{m/s}^2[/tex]Initial vertical velocity [tex]v_y = 0[/tex](a) Time in the air:The time of flight depends only on the vertical motion. Using:[tex]h = \frac{1}{2} g t^2 \implies t = \sqrt{\frac{2h}{g}}[/tex]Let's assume the cliff height is [tex]h = 20 \, \text{m}[/tex] (you can replace it with the actual height if known):[tex]t = \sqrt{\frac{2(20)}{9.8}} \approx 2.02 \, \text{seconds}[/tex](b) Horizontal distance:The horizontal motion is uniform:[tex]d = v_x \cdot t[/tex]Substituting:[tex]d = 5.0 \cdot 2.02 \approx 10.1 \, \text{meters}[/tex](c) Final velocity:The final velocity has both horizontal [tex](\( v_x \))[/tex] and vertical [tex](\( v_y \))[/tex] components:1. Vertical velocity: [tex]v_y = g \cdot t \implies v_y = 9.8 \cdot 2.02 \approx 19.8 \, \text{m/s}[/tex]2. Combine [tex]v_x[/tex] and [tex]v_y[/tex] using the Pythagorean theorem: [tex]v = \sqrt{v_x^2 + v_y^2}[/tex] Substituting:[tex]v = \sqrt{(5.0)^2 + (19.8)^2} \approx \sqrt{25 + 392.04} \approx 20.4 \, \text{m/s}[/tex]
