A silver dollar is dropped from the top of a building that is 1344 feet tall. Us

Question

A silver dollar is dropped from the top of a building that is 1344 feet tall. Use the position function below for free-falling objects. s(t) = -16t^(2) + v0t + s0 (a) Determine the position and velocity functions for the coin. s(t)=_____________ v(t)=_____________ (b) Determine the average velocity on the interval [3,4]. ______________= ft/s (c) Find the instantaneous velocities when t = 3 s and t = 4 s. v(3)=________ v(4)=________ (d) Find the time required for the coin to reach the ground level. t=___________s (e) Find the velocity of the coin at impact. vf =___________ft/s

Explanation / Answer

a) position s( t ) = -16t^2 + 0 * t + 1344 = 1344 - 16t^2 from here we calculate the diatance it travellled and we find velocity from it by differentiating it Velocity is ds/dt = -32t wheb d is at 3 we solve using -16t^2 + 0 * t + 1344 b) d at 3 is -16 * 9 + 1344 = 1200 again taking d=4 d at 4 is -16 * 16 + 1344 = 1088 and findint the average velocity So average velocity is ( 1088 -1200 )/1 = -112 ft/sec now from it we find the instantaneous velocity at 3 c) Instantaneous velocity at 3 is -32*3 = -96ft/sec and same at 4 Instantaneous velocity at 4 is -32*4 = -128ft/sec now time taken to reach the ground is distance travelled to the velocity at which it has travelled d) to reach ground -16t^2 + 1344 = 0 => t^2 = 1344/16 => t = 9.165 secs after it has travelled with some accleration the final velocity before it reached the ground is e) Impact velocity is -32 * 9.165 = -293.28ft/sec

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