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= 18t and y =4t-4.9t^2, where x and y are in meters and t is in secs. a) write a vector expression for the ball's position as a function of time using the unit vectors i and j. by taking derivatives, obtain expressions for b)the velocity vector v as a function of time and c) the acceleration vector a as a function of time. d) next use the unit vector notation to write expression for the position, the velocity and the acceleration of the golf ball a t=3sec.

Explanation / Answer

Hi, Given x = 18t           y = 4t - 4.9t^2. considering the unit vectors along the x and y directions as i and j respectively, a) position of the ball at a time t = r = xi + yj = (18t) i + (4t - 4.9t^2) j m b) Velocity = V = dr/dt = (dx/dt)i + (dy/dt)j = 18 i + (4 - 9.8t) j m/s c) Acceleration = a = dV/dt = (dVx/dt)i + (dVy/dt) j = 0 i + (-9.8)j = -9.8j m/s^2. d) at a time t = 3sec,     position r = 18(3)i + (12 - 4.9(9))j = (54i -32.1j) m     velocity V = 18 i + (4 - 29.4)j = (18i - 25.2j) m/s     acceleration a = -9.8 j m/s^2 (this is independent of time). Hope this helps you. Hope this helps you.

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