PY1052 Problem Set 3 – Autumn 2004 Solutions

(5) A supply airplane diving at an angle of 35.0 ◦ with the horizontalreleases a package of supplies at an altitude of 130 m. The package hitsthe ground 3.5 s after being released. (a) What is the speed of the plane?(b) How far did the package travel horizontally during its flight?We have the following two sets of equations for the package’s motion in the x andy directions (we take the initial position of the package when it is released to bex = 0, y = 0, and will take +y to be upward and +x to be in the horizontal directionof motion of the plane):x = x 0 + v 0x t −→ x = v cos(35) m/stv x = v 0x −→ v x = v cos(35) m/sy = y 0 + v 0y t − 1 2 gt2 −→ y = −v sin(35)t − 1 2 gt2v y = v 0y − gt −→ v y = −v sin(35) − gt(a) We know that y = −130 m when t = 3.5 s. We wish to find v – we can do thisusing the equation for y:y = −v sin(35)t − 1 2 gt2y f = −v sin(35)t f − 1 2 gt2 fv sin(35)t f = −y f − 1 2 gt2 fv = −y f − 1 2 gt2 fsin(35)t f= −(−130 m) − 1 2 (9.8 m/s2 )(3.5 s) 2sin(35)(3.5 s)v = 34.8 m/s(b) Knowing v and t f , we can also find the horizontal distance travelled before thepackage hit the ground:x f = v cos(35) m/st = (34.8 m/s) cos(35)(3.5 s)x f = 99.8 m(6) The Sun has a radius of 6.96 × 10 8 m and the material at its equatorrotates about its axis once every 26 days. What is the linear velocity ofmaterial at the Sun’s equator due to its rotation?The linear velocity will be the equal tov = 2πRT= 2π(6.96 × 108 m)26 dWe must simply express the period T in seconds to get the speed in m/s:T = 26 d ×24 hrsdv = 2π(6.96 × 108 m)2.25 × 10 6 s× 3600 shr= 2.25 × 10 6 s= 1.94 × 10 3 m/s = 1.94 km/s