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 = 19.3t and y = 3.60t ? 5.09t2,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 and . (Give the answer in terms of t.)
= 1



By taking derivatives, do the following. (Give the answers in terms of t.)
(b) obtain the expression for the velocity vector as a function of time
= 2


(c) obtain the expression for the acceleration vector as a function of time
= 3


(d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 3.06 s. = 4
m
= 5
m/s
= 6
m/s2

Explanation / Answer

The horizontal motion of the ball is at cosntant speed using equation of x = Vx*t Given x = 18*t We conclude that Vx = 18 m/s Comparing the given y(t) equation of y(t) = 4.5* - 4.9*t^2 with the equation of free fall of h(t) = h0 + Vy*t + (1/2)*g*t^2 We see that h0=0 Vy=4.5 m/s (Vy>0) a=-4.9 (a

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