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=4.20t-4.90t2, where x and y in meters and t is in seconds. 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 leftarrow= m By taking derivatives, do the following. (Give the answers in terms of t.) obtain the expression for the velocity vector v leftarrow as a function of time v leftarrow= m/s obtain the expression for the acceleration vector a lefarrow as a function of time. a leftarrow= m.s2 Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t=3.18s. r lefarrow= m v leftarrow= m/s2 leftarrow= m/s2

Explanation / Answer

a)

r = 19.8 t i^ + (4.20 t - 4.90 t2) j^

b)

taking derivative of "r" relative to 't'

dr/dt = 19.8 i^ + (4.20 - 9.8 t) j^

V = 19.8 i^ + (4.20 - 9.8 t) j^

c)

taking derivative of "V" relative to "t"

dV/dt = 0 i^ + (0 - 9.8) j^

a = 0 i^ - 9.8 j^

d)

at t = 3.18 sec

r = 19.8 t i^ + (4.20 t - 4.90 t2) j^= (19.8 (3.18)) i^ + (4.20 x 3.18 - 4.90 (3.18)2)j^

r = 62.96 i^ - 36.2 j^

V = 19.8 i^ + (4.20 - 9.8 t) j^

V = 19.8 i^ + (4.20 - 9.8) j^

V = 19.8 i^ - 5.6 j^

a = 0 i^ - 9.8 j^

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