Question

Problem #1: (Projectile Motion) A drone carrying a package to be delivered to customers. The package is to be release at the right moment to hit at specific location on the customers property, A. The drone is flying horizontally at an altitude of 100 m and velocity of 120 km/h A) At what angle 0 with the horizontal should the package be released to hit A? (hint find the horizontal distance between the drone and point A) B) What is the velocity of the package when it lands at A? v=120 km/h Pah of package 100 m Newton's Law: °(a)a mim2 xx +(v)t F= G r2 Curvilinear Normal & Tangential Coordinates W =mg R2 g = go (R+h) enve Kinematics of a Particle: Circular n-t As Vavg At v=re Av aavg re2ve At a = ds dt Curvilinear Polar Coordinates: d2s dv a =s dt dt2 =re, +ree vdy= ads |a (rr2)e, + (re+ 2re)e Rectilinear Motion: Circular Polar vvo +act V=re 1 s sovot -re a,= 2v2a(s s.) agre Curvilinear Projectile Motion: Relative Motion Fr=FFAB vy(vy)-gt yy(,)tgt v(vy)-2gyyo) Law of Cosines b2c-2bc cos A a2 b2 = a2 +c2 -2ac cos B c2=a2+b2-2ab cos C Quadratic Equation ax2bxc0 -bt vb2-4ac x= 2a Units 1 km 1,000 m 1 mile 5,280 ft Constants m23 G=66.73 x 10-12 kg-s g=9.81 m/s g=32.2 ft/s 6,371 3,959 Radius of Earth km miles

Answer 1

Vo120 kmlS Imm A Con 30 Ver initial vele city 120 Kml3 120x 33.33ms in direaton Vtinal velecity yE dral velreitj in y c-g msay0 ih direction diseetion 2 2 -2x9.81X-1rD y244.294 mls 44.294 4.81 4.515S KEVoxttxuxt 33.33x4.1 150 492 m VyCVox 38.33 mls A) estan 1 Teacher's Signature. 33.6030 S0,492 (B) V- IS0133-33- \54.136 mS