A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as

Question

A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x = 19.8t and y = 4.16t - 4.90t^2, where x and y are in meters and t is in seconds. (a) Write a vector expression for the ball's position as a function of time, using the unit vectors i and j. (Give the answer in terms of t.) r = m By taking derivatives, do the following. (Give the answers in terms of t.) (b) obtain the expression for the velocity vector v as a function of time v = m/s (c) obtain the expression for the acceleration vector a as a function of time a = m/s^2 (d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 3.27 s. r = m v = m/s a = m/s^2

Explanation / Answer

(a) The vector for the ball’s position is r = 19.8ti + (4.16t – 4.90t2)j

(b)

r = 19.8ti + (4.16t – 4.90t2)j

v = dr/dt

v = 19.8ti + (4.16t – 9.80t)j

(c)

v = 19.8ti + (4.16t – 9.80t)j

a = dv/dt

a = -9.8j

(d)

t = 3. s

r = 19.8(3.27)i + (4.16(3.27) – 4.90(3.27)2)j =64.75 mi – 38.80 mj

v = 19.8i + (4.16 – 9.8(3.27))j = 19.8 m/si -27.88 m/sj

a = -9.80 m/s2j

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