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 = 16.2t and y = 3.60t - 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.) 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 (c) obtain the expression for the acceleration vector a as a function of time (d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 2.73 s.

Explanation / Answer

(a) To wrire the position vector
r = X i + y j
r = 16.2t i + (3.6t - 4.9t2) j (where i and j represent the unit vector in x and y direction )
(b) Velocity vector
V = dr/dt
V = 16.2 i + (3.6 - (2*4.9)t) j
V = 16.2 i + (3.6 - 9.8t)j
(c) accelearation = dV/dt
a = 0 i + (0 - 9.8) j
a = -9.8 j
that means acceleration is independent of time.
(d) at t = 2.73 s
r = 16.2*2.73 i + (3.6 *2.73 - (4.9*2.732))j
r = 44.226 i - 26.69 j m
magnitude of r = (x2 + y2)1/2 = 51.826 m

Velocity
V = 16.2 i + (3.6 - 9.8t)j at t = 2.73
V = 16.2 i + (3.6 - 9.8*2.73) j
V = 16.2 i - 23.154 j
V = 28.25 m/s
Acceleration
a = -9.8 j which is independent of time
so its magnitude is
a = 9.8 m/s2

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